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Questions: Algebra BusinessCalculus
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Evaluate \(\displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}}\)
This is an indeterminate form of the type \(\displaystyle 1^\infty\). Taking the natural logarithm of both sides gives: \begin{equation*} \ln(L) = \ln\left( \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} \right) \end{equation*}Pulling the limit out of the continuous function and using log properties gives: \begin{equation*} \ln(L) = \lim_{x \to \infty}\frac{3 x}{2}\ln\left(1 + \frac{3}{x} \right) \end{equation*}This is an indeterminate form of the type \(\displaystyle 0 \cdot \infty\). Converting it to type \(\displaystyle \frac{0}{0}\) and using L'Hospitials rule gives: \begin{equation*} \ln(L) = \lim_{x \to \infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{2}{3 x}} = \frac{- \frac{3}{x^{2}}}{- \frac{2}{3 x^{2}}} = \frac{9}{2} \end{equation*}Solving for \(\displaystyle L\) gives \(\displaystyle L = e^{\frac{9}{2}}\)
\begin{question}Evaluate $\lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}}$
\soln{9cm}{This is an indeterminate form of the type $1^\infty$. Taking the natural logarithm of both sides gives:
\begin{equation*} \ln(L) = \ln\left( \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} \right) \end{equation*}Pulling the limit out of the continuous function and using log properties gives:
\begin{equation*} \ln(L) = \lim_{x \to \infty}\frac{3 x}{2}\ln\left(1 + \frac{3}{x} \right) \end{equation*}This is an indeterminate form of the type $0 \cdot \infty$. Converting it to type $\frac{0}{0}$ and using L'Hospitials rule gives:
\begin{equation*} \ln(L) = \lim_{x \to \infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{2}{3 x}} = \frac{- \frac{3}{x^{2}}}{- \frac{2}{3 x^{2}}} = \frac{9}{2} \end{equation*}Solving for $L$ gives $L = e^{\frac{9}{2}}$}
\end{question}
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\begin{document}\begin{question}(10pts) The question goes here!
\soln{9cm}{The solution goes here.}
\end{question}\end{document}<p> <p>Evaluate <img class="equation_image" title=" \displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} " src="/equation_images/%20%5Cdisplaystyle%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cleft%281%20%2B%20%5Cfrac%7B3%7D%7Bx%7D%5Cright%29%5E%7B%5Cfrac%7B3%20x%7D%7B2%7D%7D%20" alt="LaTeX: \displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} " data-equation-content=" \displaystyle \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} " /> </p> </p><p> <p>This is an indeterminate form of the type <img class="equation_image" title=" \displaystyle 1^\infty " src="/equation_images/%20%5Cdisplaystyle%201%5E%5Cinfty%20" alt="LaTeX: \displaystyle 1^\infty " data-equation-content=" \displaystyle 1^\infty " /> . Taking the natural logarithm of both sides gives:
<img class="equation_image" title=" \ln(L) = \ln\left( \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} \right) " src="/equation_images/%20%20%5Cln%28L%29%20%3D%20%5Cln%5Cleft%28%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%20%5Cleft%281%20%2B%20%5Cfrac%7B3%7D%7Bx%7D%5Cright%29%5E%7B%5Cfrac%7B3%20x%7D%7B2%7D%7D%20%5Cright%29%20%20" alt="LaTeX: \ln(L) = \ln\left( \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} \right) " data-equation-content=" \ln(L) = \ln\left( \lim_{x \to \infty} \left(1 + \frac{3}{x}\right)^{\frac{3 x}{2}} \right) " /> Pulling the limit out of the continuous function and using log properties gives:
<img class="equation_image" title=" \ln(L) = \lim_{x \to \infty}\frac{3 x}{2}\ln\left(1 + \frac{3}{x} \right) " src="/equation_images/%20%20%5Cln%28L%29%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B3%20x%7D%7B2%7D%5Cln%5Cleft%281%20%2B%20%5Cfrac%7B3%7D%7Bx%7D%20%5Cright%29%20%20" alt="LaTeX: \ln(L) = \lim_{x \to \infty}\frac{3 x}{2}\ln\left(1 + \frac{3}{x} \right) " data-equation-content=" \ln(L) = \lim_{x \to \infty}\frac{3 x}{2}\ln\left(1 + \frac{3}{x} \right) " /> This is an indeterminate form of the type <img class="equation_image" title=" \displaystyle 0 \cdot \infty " src="/equation_images/%20%5Cdisplaystyle%200%20%5Ccdot%20%5Cinfty%20" alt="LaTeX: \displaystyle 0 \cdot \infty " data-equation-content=" \displaystyle 0 \cdot \infty " /> . Converting it to type <img class="equation_image" title=" \displaystyle \frac{0}{0} " src="/equation_images/%20%5Cdisplaystyle%20%5Cfrac%7B0%7D%7B0%7D%20" alt="LaTeX: \displaystyle \frac{0}{0} " data-equation-content=" \displaystyle \frac{0}{0} " /> and using L'Hospitials rule gives:
<img class="equation_image" title=" \ln(L) = \lim_{x \to \infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{2}{3 x}} = \frac{- \frac{3}{x^{2}}}{- \frac{2}{3 x^{2}}} = \frac{9}{2} " src="/equation_images/%20%20%5Cln%28L%29%20%3D%20%5Clim_%7Bx%20%5Cto%20%5Cinfty%7D%5Cfrac%7B%5Cln%5Cleft%281%20%2B%20%5Cfrac%7B3%7D%7Bx%7D%5Cright%29%7D%7B%5Cfrac%7B2%7D%7B3%20x%7D%7D%20%3D%20%5Cfrac%7B-%20%5Cfrac%7B3%7D%7Bx%5E%7B2%7D%7D%7D%7B-%20%5Cfrac%7B2%7D%7B3%20x%5E%7B2%7D%7D%7D%20%3D%20%5Cfrac%7B9%7D%7B2%7D%20%20" alt="LaTeX: \ln(L) = \lim_{x \to \infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{2}{3 x}} = \frac{- \frac{3}{x^{2}}}{- \frac{2}{3 x^{2}}} = \frac{9}{2} " data-equation-content=" \ln(L) = \lim_{x \to \infty}\frac{\ln\left(1 + \frac{3}{x}\right)}{\frac{2}{3 x}} = \frac{- \frac{3}{x^{2}}}{- \frac{2}{3 x^{2}}} = \frac{9}{2} " /> Solving for <img class="equation_image" title=" \displaystyle L " src="/equation_images/%20%5Cdisplaystyle%20L%20" alt="LaTeX: \displaystyle L " data-equation-content=" \displaystyle L " /> gives <img class="equation_image" title=" \displaystyle L = e^{\frac{9}{2}} " src="/equation_images/%20%5Cdisplaystyle%20L%20%3D%20e%5E%7B%5Cfrac%7B9%7D%7B2%7D%7D%20" alt="LaTeX: \displaystyle L = e^{\frac{9}{2}} " data-equation-content=" \displaystyle L = e^{\frac{9}{2}} " /> </p> </p>