\hfill \hfill } %} \newtheorem{theorem}{Theorem} \newtheorem{acknowledgement}[theorem]{Acknowledgement} \newtheorem{algorithm}[theorem]{Algorithm} \newtheorem{axiom}[theorem]{Axiom} \newtheorem{case}[theorem]{Case} \newtheorem{claim}[theorem]{Claim} \newtheorem{conclusion}[theorem]{Conclusion} \newtheorem{condition}[theorem]{Condition} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{criterion}[theorem]{Criterion} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{exercise}{Exercise} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{notation}[theorem]{Notation} \newtheorem{problem}[theorem]{Problem} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{remark}[theorem]{Remark} \newtheorem{solution}[theorem]{Solution} \newtheorem{summary}[theorem]{Summary} \newenvironment{proof}[1][Proof]{\noindent\textbf{#1.} }{\ \rule{0.5em}{0.5em}} \input{tcilatex} \geometry{left=0.4in,right=0.5in,top=0.5in,bottom=0.4in} \pagestyle{fancy} \lhead{\color{blue} \Large Lecture Notes} \chead{\color{black} \LARGE Limits at Infinity - Part 2} \rhead{\large page \ \thepage} \cfoot{} \lfoot{\small \copyright $\;$ copyright Hidegkuti, Powell, 2010} \rfoot{\small Last revised: July 6, 2015} \textwidth 7.5in \textheight 9.6in \setlength{\headheight}{30pt} \setlength{\parindent}{0in} \begin{document} \begin{center} {\LARGE Sample Problems}\bigskip \end{center} \begin{enumerate} \item Compute each of the following limits. \ Show all steps, using correct notation.\medskip %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion a) \ $\ \lim\limits_{x\rightarrow \infty }\dfrac{1}{x}$\medskip b) \ $\ \lim\limits_{x\rightarrow -\infty }\dfrac{1}{x}$\medskip c) $\ \lim\limits_{x\rightarrow \infty }\dfrac{-5}{2x^{3}}$\medskip d) \ $\lim\limits_{x\rightarrow -\infty }\left( \dfrac{-5}{2x^{3}}-7+\dfrac{8% }{x}\right) $\medskip e) \ $\lim\limits_{x\rightarrow \infty }\left( -2x^{3}+1-\dfrac{5}{x}+\dfrac{% 12}{x^{4}}\right) $\medskip f) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{3x-2}{x}$\medskip g) \ $\lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{2}}$\medskip h) \ $\lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{3}}$\medskip i) \ \ $\lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{4}}$% \medskip\ \ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \item (Polynomials) \ Compute each of the following limits. \ Show all steps, using correct notation.\medskip a) $\ \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) $ \ \ \ \ \ \ \ \medskip\ \ \ \ \ \ c) \ $% \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}+8x^{6}\right) $ b) $\ \lim\limits_{x\rightarrow \infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) $ \ \ \ \ \ \medskip\ \ \ \ \ \ \ \ \ d) \ $% \lim\limits_{x\rightarrow \infty }\left( -2x^{5}+8x^{6}\right) $ \item (Rational Functions) \ Compute each of the following limits. \ Show all steps, using correct notation.\medskip a) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{x+x^{2}-6}{6x+5x^{2}+2x^{3}}$ \ \ \ \ \ \ \ \ \ \ b) \ $\lim\limits_{x\rightarrow \infty }\dfrac{x^{2}+9}{% 5x+2x^{2}-3}$ \ \ \ \ \ \ \ \ \ \ c) \ $\lim\limits_{x\rightarrow -\infty }% \dfrac{x^{3}-9x+1}{3x^{2}-2x-15}$ \item (More indeterminates) \ Compute each of the following limits. \ Show all steps, using correct notation.\medskip %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion a) $\ \lim\limits_{x\rightarrow \infty }\left( 3^{x+1}-3^{x}\right) $\bigskip b) $\ \lim\limits_{x\rightarrow \infty }\dfrac{3\sqrt{x}+2}{5\sqrt{x}+1}$% \bigskip c) $\ \lim\limits_{x\rightarrow \infty }\left( \sqrt{2x-1}-\sqrt{2x}\right) $% \bigskip d) $\ \lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{2x}% }$\bigskip e) $\ \lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{5}-\dfrac{1}{5-% \dfrac{1}{x}}\right) $\bigskip f) $\ \lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}% }$\bigskip g) $\ \lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$% \bigskip h) $\ \lim\limits_{x\rightarrow -\infty }\left( \sqrt{3x-1}-\sqrt{3x+1}% \right) $\bigskip\ \ i) \ $\lim\limits_{x\rightarrow \infty }\left( \log _{3}4x-\log _{3}12x\right) $ j) \ \ $\lim\limits_{x\rightarrow \infty }\dfrac{\log _{3}\left( 9x^{2}\right) }{\log _{3}\left( 27x\right) }$ \ %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \end{enumerate} \pagebreak \begin{center} {\LARGE Practice Problems}\bigskip \end{center} Compute each of the following limits. \ Show all steps, using correct notation.\bigskip %TCIMACRO{\TeXButton{3col begin}{\begin{multicols}{3}}}% %BeginExpansion \begin{multicols}{3}% %EndExpansion \begin{enumerate} \item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{3x-1}}{5^{x-1}}$\bigskip \item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{3x-1}}{5^{x-1}}$\bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}}{5^{x-1}}$\bigskip \item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+3}}{5^{x-1}}$\bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}}{4^{x-1}}$\bigskip \item $\lim\limits_{x\rightarrow -\infty }\dfrac{2^{2x+3}}{4^{x-1}}$\bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x+3}\cdot 3^{x-1}}{7^{x-2}% }$\bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{2^{2x+3}\cdot 3^{x-1}}{% 7^{x-2}}$\bigskip \item $\lim\limits_{x\rightarrow \infty }\left( \sqrt{x+1}-\sqrt{x}\right) $% \bigskip \item $\lim\limits_{x\rightarrow \infty }\left( \dfrac{1}{\sqrt{x+1}}-\dfrac{% 1}{\sqrt{x}}\right) $\bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{x}{\sqrt{x-1}+\sqrt{x+1}}$% \bigskip \item $\lim\limits_{x\rightarrow -\infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{% 2x}}$\bigskip \item $\lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{a}-\dfrac{1}{a-% \dfrac{1}{x}}\right) $\bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{0.5^{x}+0.5^{-x}}{% 0.5^{x}-0.5^{-x}}$\bigskip \item $\lim\limits_{x\rightarrow -\infty }\dfrac{0.5^{x}+0.5^{-x}}{% 0.5^{x}-0.5^{-x}}$\bigskip \item $\lim\limits_{x\rightarrow \infty }\left( 2^{x+2}-2^{x}\right) $% \bigskip \item $\lim\limits_{x\rightarrow \infty }\dfrac{\log _{2}4x}{\log _{2}16x}$% \bigskip \item $\lim\limits_{x\rightarrow \infty }\left( \log _{2}4x-\log _{2}16x\right) $\bigskip \item $\lim\limits_{x\rightarrow \infty }\left( \sqrt{2x}-\sqrt{x}\right) $% \bigskip \item $\lim\limits_{x\rightarrow \infty }\left( 5^{x+2}-5^{x}\right) $% \bigskip \item $\lim\limits_{x\rightarrow \infty }\left( \dfrac{5^{x}}{5^{x+2}}% \right) $\bigskip \end{enumerate} %TCIMACRO{\TeXButton{multicol end}{\end{multicols}}}% %BeginExpansion \end{multicols}% %EndExpansion \pagebreak \begin{center} {\LARGE Sample Problems - Answers}\bigskip \end{center} \bigskip 1.) \ a) \ $\ 0$\medskip\ \ \ \ \ \ \ \ b) \ $\ 0$\ \ \ \ \ \ c) $\ 0$ \ \ \ \ d) \ $-7$ \ \ \ \ \ e) \ $-\infty $ \ \ \ \ f) \ $3$\ \ \ \ \ \ g) \ $% -\infty $\ \ \ \ \ \ h) \ $-5$\ \ \ \ \ \ i) \ \ $0$\bigskip 2.) \ a) $\ \infty $ \ \ \ \ \ \ b) $\ -\infty $\ \ \ \ \ \ \ c) \ $\infty $ \ \ \ \ \ \ d) \ $\infty \qquad \qquad $3.) \ a) \ $0$ \ \ \ \ \ \ \ \ b) \ $% \dfrac{1}{2}$\ \ \ \ \ \ \ \ \ c) \ $-\infty $\bigskip 4.) \ a) \ $\infty \qquad $b) $\ \dfrac{3}{5}\qquad $c) \ $0\qquad $d) \ $-1-% \sqrt{2}\qquad $e) \ $-\dfrac{1}{25}\qquad $f) \ $1\qquad $g) \ $1\qquad $% h)\ $\func{undefined}$ \ \ \ \bigskip \ \ \ \ \ \ i) \ $-1$ \ \ j) \ $2$\bigskip \bigskip \bigskip \begin{center} {\LARGE Practice Problems - Answers}\bigskip \bigskip \end{center} 1.) $\ \infty $ \ \ \ \ \ \ \ 2.) $\ 0$ \ \ \ \ \ 3.) $\ 0$ \ \ \ \ \ \ 4.) $% \ \infty $ \ \ \ \ \ 5.) $\ 32$ \ \ \ \ \ 6.) $\ 32$ \ \ \ \ \ \ 7. ) $\ 0$ \ \ \ \ \ \ 8.) $\ \infty $ \ \ \ \ \ 9.) $\ 0$\bigskip 10.) \ $0$ \ \ \ \ \ 11.) $\ \infty $ \ \ \ \ \ \ 12.) $\ \func{undefined}$ \ \ \ 13.) \ $-\dfrac{1}{a^{2}}$ \ \ \ \ \ 14.) \ $-1$ \ \ \ \ \ 15.) \ $1$ \ \ \ \ \ 16.) \ $\infty $ \ \ \ \ 17.) \ $1$\bigskip 18.) \ $-2$\ \ \ \ \ 19.) \ $\infty $ \ \ \ \ \ 20.) \ $\infty $ \ \ \ \ \ 21.) \ $\dfrac{1}{25}$\bigskip \bigskip \bigskip \begin{center} {\LARGE Sample Problems - Solutions}\bigskip \bigskip \end{center} \begin{enumerate} \item Compute each of the following limits. a) \ $\ \lim\limits_{x\rightarrow \infty }\dfrac{1}{x}$\medskip \newline Solution: \ This is a very important limit. \ Since the limit we are asked for is as $x$ approaches infinity, we should think of $x$ as a very large positive number. \ The reciprocal of a very large positive number is a very small positive number. \ This limit is $0$.\medskip \medskip b) \ $\ \lim\limits_{x\rightarrow -\infty }\dfrac{1}{x}$\medskip \newline Solution: \ Since the limit we are asked for is as $x$ approaches negative infinity, we should think of $x$ as a very large negative number. \ The reciprocal of a very large negative number is a very small negative number. \ This limit is $0$.\medskip \medskip c) $\ \lim\limits_{x\rightarrow \infty }\dfrac{-5}{2x^{3}}$\medskip \newline Solution: \ Since the limit we are asked for is as $x$ approaches infinity, we should think of $x$ as a very large positive number. \ We divide $-5$ by a very large positive number. \ This limit is $0$.\medskip \medskip \pagebreak d) \ $\lim\limits_{x\rightarrow -\infty }\left( \dfrac{-5}{2x^{3}}-7+\dfrac{8% }{x}\right) $\medskip \newline Solution: \ This limit is $-7$ since the other two terms aproach zero as $x$ approaches negative infinity. Using mathematical notation,% \begin{equation*} \lim\limits_{x\rightarrow -\infty }\dfrac{-5}{2x^{3}}-7+\dfrac{8}{x}% =\lim\limits_{x\rightarrow -\infty }\dfrac{-5}{2x^{3}}+\lim\limits_{x% \rightarrow -\infty }-7+\lim\limits_{x\rightarrow -\infty }\dfrac{8}{x}% =0-7+0=-7 \end{equation*} e) \ $\lim\limits_{x\rightarrow \infty }\left( -2x^{3}+1-\dfrac{5}{x}+\dfrac{% 12}{x^{4}}\right) $\medskip \newline Solution: \ This limit is $-\infty $ since the first term approaches negative infinity, the second term approaches $1$ and the other two terms aproach zero as $x$ approaches infinity. Using mathematical notation,% \begin{equation*} \lim\limits_{x\rightarrow \infty }\left( -2x^{3}+1-\dfrac{5}{x}+\dfrac{12}{% x^{4}}\right) =\lim\limits_{x\rightarrow \infty }\left( -2x^{3}\right) +\lim\limits_{x\rightarrow \infty }1+\lim\limits_{x\rightarrow \infty }\left( -\dfrac{5}{x}\right) +\lim\limits_{x\rightarrow \infty }\left( \dfrac{12}{x^{4}}\right) =-\infty +1+0+0=-\infty \end{equation*} f) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{3x-2}{x}$\medskip \newline Solution: \ This problem is similar to the previous problems after a bit of algebra. \ We simply divide by $x$ and then the limit becomes familiar.% \begin{equation*} \lim\limits_{x\rightarrow -\infty }\dfrac{3x-2}{x}=\lim\limits_{x\rightarrow -\infty }\left( \dfrac{3x}{x}-\dfrac{2}{x}\right) =\lim\limits_{x\rightarrow -\infty }\left( 3-\dfrac{2}{x}\right) =3 \end{equation*} g) \ $\lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{2}}$\medskip \newline Solution: \ \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{2}} &=&\lim\limits_{x\rightarrow \infty }\left( \dfrac{-5x^{3}}{x^{2}}+\dfrac{-2x% }{x^{2}}+\dfrac{4}{x^{2}}\right) =\lim\limits_{x\rightarrow \infty }\left( -5x-\dfrac{2}{x}+\dfrac{4}{x^{2}}\right) \\ &=&\lim\limits_{x\rightarrow \infty }\left( -5x\right) +\lim\limits_{x\rightarrow \infty }\left( -\dfrac{2}{x}\right) +\lim\limits_{x\rightarrow \infty }\left( \dfrac{4}{x^{2}}\right) =-\infty +0+0=-\infty \end{eqnarray*} h) \ $\lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{3}}$\medskip \newline Solution: \ \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{3}}% =\lim\limits_{x\rightarrow \infty }\left( \dfrac{-5x^{3}}{x^{3}}+\dfrac{-2x}{% x^{3}}+\dfrac{4}{x^{3}}\right) =\lim\limits_{x\rightarrow \infty }\left( -5-% \dfrac{2}{x^{2}}+\dfrac{4}{x^{3}}\right) =-5 \end{equation*} i) \ \ $\lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{4}}$% \medskip \newline Solution: \ \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{-5x^{3}-2x+4}{x^{4}}% =\lim\limits_{x\rightarrow \infty }\left( \dfrac{-5x^{3}}{x^{4}}+\dfrac{-2x}{% x^{4}}+\dfrac{4}{x^{4}}\right) =\lim\limits_{x\rightarrow \infty }\left( \dfrac{-5}{x}-\dfrac{2}{x^{3}}+\dfrac{4}{x^{4}}\right) =0 \end{equation*}% \pagebreak \medskip \item Compute each of the following limits. a) $\ \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) $\medskip \newline Solution: \ The first term, $-2x^{5}$ approaches infinity and the secomd term, $-8x^{4}$ approaches negative infinity. \ This does not give us enough information about the entire polynomial. \ A limit like this is called an \textbf{indeterminate}. \ We will bring this expression to a\ form that is not an indeterminate. \ In this case, factoring out the first term does the trick. \textbf{In case of a polynomial, the limits at infinity and negative infinity are }\newline \textbf{completely determined by its leading term.} \ Recall that the leading term is the highest degree term.% \begin{equation*} \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) =\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) \end{equation*}% Here is the computation showing why this is true. \ We first factor out the entire leading term.% \begin{eqnarray*} \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) &=&\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) \left( 1+\dfrac{4% }{x}-\dfrac{7}{x^{2}}+\dfrac{5}{x^{5}}\right) \\ &=&\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) \cdot \lim\limits_{x\rightarrow -\infty }\left( 1+\dfrac{4}{x}-\dfrac{7}{x^{2}}+% \dfrac{5}{x^{5}}\right) \\ &=&\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) \cdot 1=\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}\right) \end{eqnarray*}% We can now easily determine that this limit is $\infty $. \medskip \medskip b) $\ \lim\limits_{x\rightarrow \infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) $\medskip \newline Solution: \ \ \textbf{In case of a polynomial, the limits at infinity and negative infinity are }\newline \textbf{completely determined by its leading term.} \ Recall that the leading term is the highest degree term.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) =\lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) \end{equation*}% Here is the computation showing why this is true. \ We first factor out the entire leading term.% \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\left( -2x^{5}-8x^{4}+7x^{3}-10\right) &=&\lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) \left( 1+\dfrac{4% }{x}-\dfrac{7}{2x^{2}}+\dfrac{5}{x^{5}}\right) \\ &=&\lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) \cdot \lim\limits_{x\rightarrow \infty }\left( 1+\dfrac{4}{x}-\dfrac{7}{2x^{2}}+% \dfrac{5}{x^{5}}\right) \\ &=&\lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) \cdot 1=\lim\limits_{x\rightarrow \infty }\left( -2x^{5}\right) \end{eqnarray*}% We can now easily determine that this limit is $-\infty $.\medskip \medskip c) \ $\lim\limits_{x\rightarrow -\infty }\left( -2x^{5}+8x^{6}\right) $% \medskip \newline Solution: \ \textbf{In case of a polynomial, the limits at infinity and negative infinity are }\newline \textbf{completely determined by its leading term.}% \begin{eqnarray*} \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}+8x^{6}\right) &=&\lim\limits_{x\rightarrow -\infty }8x^{6}=\infty \text{ \ \ because} \\ \lim\limits_{x\rightarrow -\infty }\left( -2x^{5}+8x^{6}\right) &=&\lim\limits_{x\rightarrow -\infty }\left( 8x^{6}-2x^{5}\right) =\lim\limits_{x\rightarrow -\infty }\left( 8x^{6}\right) \left( 1-\dfrac{1}{% 4x}\right) =\lim\limits_{x\rightarrow -\infty }\left( 8x^{6}\right) \cdot \lim\limits_{x\rightarrow -\infty }\left( 1-\dfrac{1}{4x}\right) \\ &=&\lim\limits_{x\rightarrow -\infty }\left( 8x^{6}\right) \cdot 1=\lim\limits_{x\rightarrow -\infty }8x^{6} \end{eqnarray*}% We can now easily determine that this limit is $\infty $. \medskip \medskip d) \ $\lim\limits_{x\rightarrow \infty }\left( -2x^{5}+8x^{6}\right) $% \medskip \newline Solution: \ \textbf{In case of a polynomial, the limits at infinity and negative infinity are }\newline \textbf{completely determined by its leading term.}% \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\left( -2x^{5}+8x^{6}\right) &=&\lim\limits_{x\rightarrow \infty }8x^{6}=\infty \text{ \ \ because} \\ \lim\limits_{x\rightarrow \infty }\left( -2x^{5}+8x^{6}\right) &=&\lim\limits_{x\rightarrow \infty }\left( 8x^{6}-2x^{5}\right) =\lim\limits_{x\rightarrow \infty }\left( 8x^{6}\right) \left( 1-\dfrac{1}{4x% }\right) =\lim\limits_{x\rightarrow \infty }\left( 8x^{6}\right) \cdot \lim\limits_{x\rightarrow \infty }\left( 1-\dfrac{1}{4x}\right) \\ &=&\lim\limits_{x\rightarrow \infty }\left( 8x^{6}\right) \cdot 1=\infty \end{eqnarray*}% We can now easily determine that this limit is $\infty $. \medskip \medskip \item Compute each of the following limits. a) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{x+x^{2}-6}{6x+5x^{2}+2x^{3}}$% \medskip \newline Solution: \ The numerator approaches infinity and the denominator approaches negative infinity. \ This does not give us enough information about the quotient. \ A limit like this is called an \textbf{indeterminate}. \ We will bring this expression to a\ form that is not an indeterminate. Let us rearrange the polynomials in the rational function given. \ Then we will factor out the leading term in the numerator and denominator.% \begin{equation*} \lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}+x-6}{2x^{3}+5x^{2}+6x}% =\lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}\left( 1+\dfrac{1}{x}-\dfrac{% 6}{x^{2}}\right) }{2x^{3}\left( 1+\dfrac{5}{2x}+\dfrac{3}{x^{2}}\right) } \end{equation*}% We now express the limit of the product as the product of two limits% \begin{equation*} \lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}\left( 1+\dfrac{1}{x}-\dfrac{6% }{x^{2}}\right) }{2x^{3}\left( 1+\dfrac{5}{2x}+\dfrac{3}{x^{2}}\right) }% =\lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}}{2x^{3}}\cdot \lim\limits_{x\rightarrow -\infty }\dfrac{\left( 1+\dfrac{1}{x}-\dfrac{6}{% x^{2}}\right) }{\left( 1+\dfrac{5}{2x}+\dfrac{3}{x^{2}}\right) } \end{equation*}% The first expression can be simplified and thus has a limit we can easily determine its limit. \ The second expression, although looks unfriendly, is always going to approach $1$.% \begin{equation*} \lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}}{2x^{3}}\cdot \lim\limits_{x\rightarrow -\infty }\dfrac{\left( 1+\dfrac{1}{x}-\dfrac{6}{% x^{2}}\right) }{\left( 1+\dfrac{5}{2x}+\dfrac{3}{x^{2}}\right) }% =\lim\limits_{x\rightarrow -\infty }\dfrac{1}{2x}\cdot 1=0\cdot 1=0 \end{equation*}% The entire computation should look like this:% \begin{eqnarray*} \lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}+x-6}{2x^{3}+5x^{2}+6x} &=&\lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}\left( 1+\dfrac{1}{x}-% \dfrac{6}{x^{2}}\right) }{2x^{3}\left( 1+\dfrac{5}{2x}+\dfrac{3}{x^{2}}% \right) }=\lim\limits_{x\rightarrow -\infty }\dfrac{x^{2}}{2x^{3}}\cdot \lim\limits_{x\rightarrow -\infty }\left( \dfrac{1+\dfrac{1}{x}-\dfrac{6}{% x^{2}}}{1+\dfrac{5}{2x}+\dfrac{3}{x^{2}}}\right) \\ &=&\lim\limits_{x\rightarrow -\infty }\dfrac{1}{2x}\cdot 1=0\cdot 1=0 \end{eqnarray*}% \pagebreak b) \ $\lim\limits_{x\rightarrow \infty }\dfrac{x^{2}+9}{5x+2x^{2}-3}$% \medskip \newline Solution: \ \ \ Both numerator and denominator approach infinity. \ This does not give us enough information about the quotient. \ A limit like this is called an \textbf{indeterminate}. \ We will bring this expression to a\ form that is not an indeterminate. Let us rearrange the polynomials in the rational function given. \ Then we will factor out the leading term in the numerator and denominator.% \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\dfrac{x^{2}+9}{2x^{2}+5x-3} &=&\lim\limits_{x\rightarrow \infty }\dfrac{x^{2}\left( 1+\dfrac{9}{x^{2}}% \right) }{2x^{2}\left( 1+\dfrac{5}{2x}-\dfrac{3}{2x^{2}}\right) }% =\lim\limits_{x\rightarrow \infty }\dfrac{x^{2}}{2x^{2}}\cdot \lim\limits_{x\rightarrow \infty }\dfrac{1+\dfrac{9}{x^{2}}}{1+\dfrac{5}{2x}-% \dfrac{3}{2x^{2}}} \\ &=&\lim\limits_{x\rightarrow \infty }\dfrac{1}{2}\cdot \lim\limits_{x\rightarrow \infty }\dfrac{1+\dfrac{9}{x^{2}}}{1+\dfrac{5}{2x}-% \dfrac{3}{2x^{2}}}=\dfrac{1}{2}\cdot 1=\dfrac{1}{2} \end{eqnarray*} c) \ $\lim\limits_{x\rightarrow -\infty }\dfrac{x^{3}-9x+1}{3x^{2}-2x-15}$% \medskip \newline Solution: \ \ \begin{eqnarray*} \lim\limits_{x\rightarrow -\infty }\dfrac{x^{3}-9x+1}{3x^{2}-2x-15} &=&\lim\limits_{x\rightarrow -\infty }\dfrac{x^{3}\left( 1-\dfrac{9}{x^{2}}+% \dfrac{1}{x^{3}}\right) }{3x^{2}\left( 1-\dfrac{2}{3x}-\dfrac{5}{x^{2}}% \right) }=\lim\limits_{x\rightarrow -\infty }\dfrac{x^{3}}{3x^{2}}\cdot \lim\limits_{x\rightarrow -\infty }\dfrac{1-\dfrac{9}{x^{2}}+\dfrac{1}{x^{3}}% }{1-\dfrac{2}{3x}-\dfrac{5}{x^{2}}} \\ &=&\left( \lim\limits_{x\rightarrow -\infty }\dfrac{x^{3}}{3x^{2}}\right) \cdot 1=\left( \lim\limits_{x\rightarrow -\infty }\dfrac{x}{3}\right) \cdot 1=-\infty \cdot 1=-\infty \end{eqnarray*} \item (More indeterminates) \ Compute each of the following limits. \ Show all steps, using correct notation.\medskip a) \ $\lim\limits_{x\rightarrow \infty }\left( 3^{x+1}-3^{x}\right) $ Solution: \ This is an "infinity minus infinity" type of an indeterminate. \ We need to first transform this expression until it is no longer an indeterminate.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\left( 3^{x+1}-3^{x}\right) =\lim\limits_{x\rightarrow \infty }\left( 3^{x}\cdot 3-3^{x}\right) =\lim\limits_{x\rightarrow \infty }\left( 3\cdot 3^{x}-3^{x}\right) =\lim\limits_{x\rightarrow \infty }\left( 2\cdot 3^{x}\right) =2\lim\limits_{x\rightarrow \infty }3^{x}=\infty \end{equation*} b) $\ \lim\limits_{x\rightarrow \infty }\dfrac{3\sqrt{x}+2}{5\sqrt{x}+1}$ Solution: \ Since $\lim\limits_{x\rightarrow \infty }\sqrt{x}=\infty ,$ clearly $\lim\limits_{x\rightarrow \infty }\dfrac{1}{\sqrt{x}}=0.$ \ We will use this fact; we factor out $\sqrt{x}$ from both numerator and denominator.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{3\sqrt{x}+2}{5\sqrt{x}+1}% =\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}\left( 3+\dfrac{2}{\sqrt{x}% }\right) }{\sqrt{x}\left( 5+\dfrac{1}{\sqrt{x}}\right) }=\lim\limits_{x% \rightarrow \infty }\dfrac{3+\dfrac{2}{\sqrt{x}}}{5+\dfrac{1}{\sqrt{x}}}=% \dfrac{3}{5} \end{equation*} c) \ $\lim\limits_{x\rightarrow \infty }\left( \sqrt{2x-1}-\sqrt{2x}\right) $ Solution: \ We will transform this expression by multiplying it by $1$, written as a fraction with numerator and denominator both being the conjugate of the expression.% \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\left( \sqrt{2x-1}-\sqrt{2x}\right) &=&\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{2x-1}-\sqrt{2x}}{1}\cdot \dfrac{\sqrt{2x-1}+\sqrt{2x}}{\sqrt{2x-1}+\sqrt{2x}} \\ &=&\lim\limits_{x\rightarrow \infty }\dfrac{\left( 2x-1\right) -\left( 2x\right) }{\sqrt{2x-1}+\sqrt{2x}}=\lim\limits_{x\rightarrow \infty }\dfrac{% -1}{\sqrt{2x-1}+\sqrt{2x}}=0 \end{eqnarray*} d) \ $\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{2x}% }$ Solution: \ We factor out $\sqrt{x}$ \ from both numerator and denominator.% \begin{eqnarray*} \lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x+1}-\sqrt{2x}} &=&\lim\limits_{x\rightarrow \infty }\dfrac{\sqrt{x}}{\sqrt{x}\left( \dfrac{% \sqrt{x+1}}{\sqrt{x}}-\sqrt{2}\right) }=\lim\limits_{x\rightarrow \infty }% \dfrac{1}{\sqrt{\dfrac{x+1}{x}}-\sqrt{2}} \\ &=&\lim\limits_{x\rightarrow \infty }\dfrac{1}{\sqrt{1+\dfrac{1}{x}}-\sqrt{2}% }=\dfrac{1}{1-\sqrt{2}}=\dfrac{1}{1-\sqrt{2}}\cdot \dfrac{1+\sqrt{2}}{1+% \sqrt{2}}=\dfrac{1+\sqrt{2}}{-1} \\ &=&-1-\sqrt{2} \end{eqnarray*}% \bigskip e) \ $\lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{5}-\dfrac{1}{5-% \dfrac{1}{x}}\right) $ Solution: \ We just need to simplify the complex fraction. \ As it turns out, this problem boils down to a type we have already seen.% \begin{eqnarray*} x\left( \dfrac{1}{5}-\dfrac{1}{5-\dfrac{1}{x}}\right) &=&x\left( \dfrac{1}{5}% -\dfrac{1}{~~\dfrac{5x-1}{x}~~}\right) =x\left( \dfrac{1}{5}-\dfrac{x}{5x-1}% \right) =x\left( \dfrac{\left( 5x-1\right) -5x}{5\left( 5x-1\right) }\right) \\ &=&x\left( \dfrac{5x-1-5x}{5\left( 5x-1\right) }\right) =x\dfrac{-1}{25x-5}=% \dfrac{-x}{25x-5} \end{eqnarray*}% Thus% \begin{equation*} \lim\limits_{x\rightarrow \infty }x\left( \dfrac{1}{5}-\dfrac{1}{5-\dfrac{1}{% x}}\right) =\lim\limits_{x\rightarrow \infty }\dfrac{-x}{25x-5}% =\lim\limits_{x\rightarrow \infty }\dfrac{x\left( -1\right) }{x\left( 25-% \dfrac{5}{x}\right) }=-\dfrac{1}{25} \end{equation*} f) \ $\lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}% }$ Solution: \ We will factor out $x$ from both numerator and denominator, and use the fact that $\lim\limits_{x\rightarrow \infty }\dfrac{1}{x^{k}}=0$ for all positive integers $k$.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}}% =\lim\limits_{x\rightarrow \infty }\dfrac{x\left( 1+\dfrac{1}{x^{2}}\right) }{x\left( 1-\dfrac{1}{x^{2}}\right) }=\lim\limits_{x\rightarrow \infty }% \dfrac{1+\dfrac{1}{x^{2}}}{1-\dfrac{1}{x^{2}}}=1 \end{equation*} \pagebreak g) \ $\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}$ Solution: \ First, $\lim\limits_{x\rightarrow \infty }2^{x}=\infty $ (and so $2^{x}$ is large) and $\lim\limits_{x\rightarrow \infty }2^{-x}=0$ \ (and so $2^{-x}$ is small). \ With that in mind, this limit is similar to \ $% \lim\limits_{x\rightarrow \infty }\dfrac{x+\dfrac{1}{x}}{x-\dfrac{1}{x}}$. \ The solution also will be similar. \ \ We will factor out $2^{x}$ \ from both numerator and denominator.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+2^{-x}}{2^{x}-2^{-x}}% =\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}+\dfrac{1}{2^{x}}}{2^{x}-% \dfrac{1}{2^{x}}}=\lim\limits_{x\rightarrow \infty }\dfrac{2^{x}\left( 1+% \dfrac{1}{\left( 2^{x}\right) ^{2}}\right) }{2^{x}\left( 1-\dfrac{1}{\left( 2^{x}\right) ^{2}}\right) }=\lim\limits_{x\rightarrow \infty }\dfrac{1+% \dfrac{1}{2^{2x}}}{1-\dfrac{1}{2^{2x}}}=1 \end{equation*} h)\ $\ \lim\limits_{x\rightarrow -\infty }\left( \sqrt{3x-1}-\sqrt{3x+1}% \right) $ Solution: \ When $x\rightarrow -\infty ,$ then we may assume it is negative. \ Then the expressions under the square root are negative and the function is not defined. \ Thus, there is no limit at negative infinity. \ The answer is: $\func{undefined}$. i) \ $\lim\limits_{x\rightarrow \infty }\left( \log _{3}4x-\log _{3}12x\right) $ Solution: \ This is an infinity minus infinity type of an indeterminate. \ We will use properties of lgarithms to bring it to a form that is no longer an indeterminate. \ Recall the rule $\log _{3}a-\log _{3}b=\log _{3}\left( \dfrac{a}{b}\right) $% \begin{equation*} \lim\limits_{x\rightarrow \infty }\left( \log _{3}4x-\log _{3}12x\right) =\lim\limits_{x\rightarrow \infty }\left( \log _{3}\dfrac{4x}{12x}\right) =\lim\limits_{x\rightarrow \infty }\left( \log _{3}\dfrac{4}{12}\right) =\lim\limits_{x\rightarrow \infty }\left( \log _{3}\dfrac{1}{3}\right) =\lim\limits_{x\rightarrow \infty }\left( -1\right) =-1 \end{equation*} j) \ $\lim\limits_{x\rightarrow \infty }\dfrac{\log _{3}\left( 9x^{2}\right) }{\log _{3}\left( 27x\right) }$ Solution: \ This is an infinity divided by infinity type of an indeterminate. \ We will use properties of lgarithms to bring it to a form that is no longer an indeterminate. \ Recall the rules $\log _{3}a+\log _{3}b=\log _{3}ab$ \ and \ $\log _{3}\left( a^{b}\right) =b\log _{3}a$.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{\log _{3}\left( 9x^{2}\right) }{% \log _{3}\left( 27x\right) }=\lim\limits_{x\rightarrow \infty }\dfrac{\log _{3}9+\log _{3}x^{2}}{\log _{3}27+\log _{3}x}=\lim\limits_{x\rightarrow \infty }\dfrac{2+2\log _{3}x}{3+\log _{3}x} \end{equation*}% Since $\lim\limits_{x\rightarrow \infty }\log _{3}x=\infty $, this is still an indeterminate of an infinity divided by infinity type. \ However, it is very similar to $\lim\limits_{x\rightarrow \infty }\dfrac{2+2x}{3+x}$ and so we will transform it using the same technique. \ We will factor out $\log _{3}x$ from both numerator and denominator and use the fact that $% \lim\limits_{x\rightarrow \infty }\dfrac{1}{\log _{3}x}=0$.% \begin{equation*} \lim\limits_{x\rightarrow \infty }\dfrac{2+2\log _{3}x}{3+\log _{3}x}% =\lim\limits_{x\rightarrow \infty }\dfrac{\log _{3}x\left( \dfrac{2}{\log _{3}x}+2\right) }{\log _{3}x\left( \dfrac{3}{\log _{3}x}+1\right) }=\dfrac{2% }{1}=2 \end{equation*} \end{enumerate} \bigskip \bigskip {\small %TCIMACRO{\TeXButton{\small}{\small}}% %BeginExpansion \small% %EndExpansion } \href{https://teaching.martahidegkuti.com/shared/lnotes/lecturenotes.html}{% For more documents like this, visit our page at\ https://teaching.martahidegkuti.com and click on Lecture Notes. \ E-mail questions or comments to mhidegkuti@ccc.edu.} \end{document}