Is this the correct way to evaluate the limit of this function?

I need to compute the limit of $10xe^{1/x} - 10x$ as $x$ approaches infinity. Since $e^{1/x}$ approaches 1 as x approaches infinity, shouldn't the expression approach 0? I also tried doing this using l'hopital's rule but couldn't arrive at the correct answer, which is 10. Any help or hints would be appreciated.

Answer

The fact that $e^{1/x}$ approaches $1$ as $x\to\infty$ tempts us to believe that the limit is zero. But we should never rely on such hand-wavy arguments. The issue is that $e^{1/x}$ approaches one, while at the same time $10x$ approaches infinity. It is difficult to say exactly what the limit is. Your are also right that we need to use L'Hopital's rule. But L'Hopital's rule needs a fraction. So we rewrite the limit as

$$\lim_{x\to\infty} 10xe^{1/x}-10x=\lim_{x\to\infty}10x(e^{1/x}-1)=\lim_{x\to\infty}\frac{e^{1/x}-1}{\frac{1}{10x}}.$$

The limit of both numerator and denominator is $0$. So we attempt to apply L'Hopital's rule. In doing so, we get
$$\lim_{x\to\infty}\frac{-e^{1/x}/x^2}{-1/10x^2}=\lim_{x\to\infty} 10e^{1/x}=10.$$
So indeed, our intuition has failed us! Good thing we have L'Hopital's rule.