Chapter 4 - Applications of the Derivative - 4.7 L'Hopital's Rule - 4.7 Exercises - Page 307: 29

The solution $$\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=1.$$

We will calculate this limit using L'Hopital's rule. "LR" will stand for "Apply L'Hopital's rule": $$\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\left[\frac{0}{0}\right][\text{LR}]=\lim_{x\to\infty}\frac{(e^{1/x}-1)'}{(1/x)'}=\lim_{x\to\infty}\frac{e^{1/x}(1/x)'}{(1/x)'}=\lim_{x\to\infty}e^{1/x}=e^{1/\infty}=e^0=1.$$