## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^x=e^a.$$
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 $\left(1+\frac{a}{x}\right)^{x}=e^{\ln \left(1+\frac{a}{x}\right)^{x}}=e^{x\ln\left(1+\frac{a}{x}\right)}$ 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 $$L=\lim_{x\to\infty}\left(1+\frac{a}{x}\right)^{x}=\lim_{x\to\infty}e^{x\ln\left(1+\frac{a}{x}\right)}=e^{\lim_{x\to\infty}x\ln\left(1+\frac{a}{x}\right)}=e^l,$$ where we have denoted $l=\lim_{x\to\infty}x\ln\left(1+\frac{a}{x}\right).$ Now let's calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$l=\lim_{x\to\infty}x\ln\left(1+\frac{a}{x}\right)=\lim_{x\to\infty}\frac{\ln\left(1+\frac{a}{x}\right)}{\frac{1}{x}}=\left[\frac{\ln(1+a/\infty)}{\frac{1}{\infty}}\right]=\left[\frac{\ln(1+0)}{0}\right]=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{\left(\ln\left(1+\frac{a}{x}\right)\right)'}{\left(\frac{1}{x}\right)'}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{a}{x}}\left(1+\frac{a}{x}\right)'}{-\frac{1}{x^2}}=\lim_{x\to\infty}\frac{\frac{1}{1+\frac{a}{x}}\left(-\frac{a}{x^2}\right)}{-\frac{1}{x^2}}=\lim_{x\to\infty}\frac{a}{1+\frac{1}{x}}=\left[\frac{a}{1+\frac{1}{\infty}}\right]=\left[\frac{a}{1+0}\right]=a.$$ Putting this into the initial limit we have $$L=e^l=e^a.$$