Determine the $\displaystyle \lim_{x \to \infty} (e^x + x)^{\frac{1}{x}}$. Use L'Hospital's Rule where appropriate. Use some Elementary method if posible. If L'Hospitals Rule doesn't apply. Explain why.

If we let $y = (e^x + x)^{\frac{1}{x}}$, then

$ \ln y = \left( \frac{1}{x} \right) \ln (e^x + x)$

So,

$\displaystyle \lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{\ln(e^x+x)}{x}$

By applying L'Hospital's Rule...

$\displaystyle \lim_{x \to \infty} \frac{\ln(e^x+x)}{x} = \lim_{x \to \infty} \frac{\frac{e^x+1}{e^x + x}}{1} = \lim_{x \to \infty} \frac{e^x +1}{e^x + x}$

If we evaluate the limit, we will still get indeterminate form, so we next apply L'Hospital's Rule once more. Thus,

$\displaystyle \lim_{x \to \infty} \frac{e^x +1}{e^x +x} = \lim_{x \to \infty} \frac{e^x}{e^x +1}$

Again, by applying L'Hospital's Rule...

$\displaystyle \lim_{x \to \infty} \frac{e^x}{e^x + 1} = \lim_{x \to \infty} \frac{e^x}{e^x} = 1$

Hence,

$\displaystyle \lim_{x \to \infty} \ln y = \lim_{x \to \infty} \frac{\ln(e^x+x)}{x} = 1$

Therefore, we have

$\displaystyle \lim_{x \to \infty} (e^x + x)^{\frac{1}{x}} = \lim_{x \to \infty} e^{\ln y} = e^1 = e$