## Calculus: Early Transcendentals (2nd Edition)

The solution is $$\lim_{x\to0}(e^{ax}+x)^{1/x}=e^{a+1}.$$
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 $\left(e^{ax}+x\right)^{1/x}=e^{\ln \left(e^{ax}+x\right)^{1/x}}=e^{\frac{1}{x}\ln\left(e^{ax}+x\right)}$ 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 $$L=\lim_{x\to0}\left(e^{ax}+x\right)^{1/x}=\lim_{x\to0}e^{\frac{1}{x}\ln\left(e^{ax}+x\right)}=e^{\lim_{x\to0}\frac{1}{x}\ln\left(e^{ax}+x\right)}=e^l,$$ where we have denoted $l=\lim_{x\to0}\frac{1}{x}\ln\left(e^{ax}+x\right).$ Now let's calculate this limit. "LR" will stand for "Apply L'Hopital's rule". $$\lim_{x\to0}\frac{1}{x}\ln\left(e^{ax}+x\right)=\lim_{x\to0}\frac{\ln(e^{ax}+x)}{x}=\left[\frac{\ln(e^{a\cdot\infty}+\infty)}{\infty}\right]=\left[\frac{\infty}{\infty}\right][\text{LR}]=\lim_{x\to0}\frac{(\ln(e^{ax}+x))'}{(x)'}=\lim_{x\to0}\frac{\frac{1}{e^{ax}+x}(e^{ax}+x)'}{1}=\lim_{x\to0}\frac{ae^{ax}+1}{e^{ax}+x}=\frac{ae^{a\cdot0}+1}{e^{ae^\cdot0}+0}=a+1.$$ Putting this into the initial limit we get $$L=e^l=e^{a+1}.$$