## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{z\to\infty}\left(1+\frac{10}{z^2}\right)^{z^2}=e^{10}.$$
To solve this limit we will transform the expression under the limit. Since the natural exponential function and the natural logarithmic function are inverse to each other we have that $c=e^{\ln c}.$ This gives $\left(1+\frac{10}{z^2}\right)^{z^2}=e^{\ln \left(1+\frac{10}{z^2}\right)^{z^2}}=e^{z^2\ln\left(1+\frac{10}{z^2}\right)}$ where we used the logarithmic rule $\ln b^c=c\ln b$. Also because the exponential function is continuous it can exchange places with the limit so we get $$L=\lim_{z\to\infty}\left(1+\frac{10}{z^2}\right)^{z^2}=\lim_{z\to\infty}e^{z^2\ln\left(1+\frac{10}{z^2}\right)}=e^{\lim_{z\to\infty}{z^2}\ln\left(1+\frac{10}{z^2}\right)}=e^l,$$ where we have denoted $l=\lim_{z\to\infty}{z^2}\ln\left(1+\frac{10}{z^2}\right).$ Now let's calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$l=\lim_{z\to\infty}{z^2}\ln\left(1+\frac{10}{z^2}\right)=\lim_{z\to\infty}\frac{\ln\left(1+\frac{10}{z^2}\right)}{\frac{1}{z^2}}=\left[\frac{\ln(1+10/\infty^2)}{\frac{1}{\infty^2}}\right]=\left[\frac{\ln(1+0)}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{z\to\infty}\frac{\left(\ln\left(1+\frac{10}{z^2}\right)\right)'}{\left(\frac{1}{z^2}\right)'}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{10}{z^2}}\left(1+\frac{10}{z^2}\right)'}{-\frac{2}{z^3}}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{10}{z^2}}10\left(-\frac{2}{z^3}\right)}{-\frac{2}{z^3}}=\lim_{x\to\infty}\frac{10}{1+\frac{10}{z^2}}=\left[\frac{10}{1+10/\infty^2}\right]=\left[\frac{10}{1+0}\right]=10.$$ Putting this into the initial limit we have $$L=e^l=e^{10}.$$