L'Hospital's Rule for a function not a fraction My answer was $\infty$ because at the end I got $\infty-\infty$ , yet I feel this is wrong?
$$\lim_{x\to \infty}(xe^\frac{1}{x}-x)$$
 A: As an alternative note that
$$\left(x\left[\left(1+\frac1x\right)^x\right]^\frac{1}{x}-x\right)=x+1-x=1
\leq (xe^\frac{1}{x}-x)1\leq$$
$$\leq \left(x\left[\left(1+\frac1x\right)^{x+1}\right]^\frac{1}{x}-x\right)=\left(x+1\right)\left(1+\frac1x\right)^{\frac1x}-x\stackrel{\text{Bernoulli's inequality}}\leq$$
$$\leq(x+1)\left(1+\frac1{x^2}\right)-x=x+\frac1x+1+\frac1{x^2}-x=1+\frac1x+\frac1{x^2}\to1$$
thus for squeeze theorem
$$(xe^\frac{1}{x}-x)\to1$$
A: $\infty-\infty$ is called an indeterminate form. It is so called because it's not really clear what's going to with the limit as $x$ approaches a certain number or, as in your case here, infinity. For all intents and purposes, an indeterminate form can't be and never is the answer to a limit problem because it doesn't answer the question of whether the limit exists or not. When you get an indeterminate form, it means that you actually need to do some more work to find the limit. In your case, notice that $\frac{x}{1}=\frac{1}{\frac{1}{x}}$ as long as $x$ does not equal zero (it's not in our case here):
\begin{align}\require{cancel}
\lim_{x\to\infty}\left(xe^\frac{1}{x}-x\right)
&=\lim_{x\to\infty}x\left(e^\frac{1}{x}-1\right)\\
&=\lim_{x\to\infty}\frac{e^\frac{1}{x}-1}{\frac{1}{x}}\\
&=\frac{0}{0}\text{ is an indeterminate form, apply L'Hospital's rule}\\
&=\lim_{x\to\infty}\frac{\left(e^\frac{1}{x}-1\right)'}{\left(\frac{1}{x}\right)'}\\
&=\lim_{x\to\infty}\frac{-\frac{1}{x^2}e^{\frac{1}{x}}}{-\frac{1}{x^2}}\\
&=\lim_{x\to\infty}e^{\frac{1}{x}}\\
&=e^{0}\\
&=1
\end{align}
A: It's $$\lim_{x\rightarrow\infty}\frac{e^{\frac{1}{x}}-1}{\frac{1}{x}}=1$$
because for example $$\left(e^x\right)'_{x=0}=e^0=1.$$
