How can I use l'Hospital rule to calculate
$$\lim_{x\to\infty }\frac{x-\sin(x)}{e^x-1-x-\frac{x^2}{2}}$$
?
Intuitively, it seems like the denominator tends to $\infty$ faster
than the numerator, so I guess the limit is $0$. But how can I use
l'Hospital's rule to calculate this formally?
Differentiating the denominator and numerator once yields $\lim_{x\to\infty }\frac{1-\cos(x)}{e^x-1-x}$, and the numerator does not even have a limit in this case. Dividing everything by $x$ or $x^2$ doesn't seem to help in anything. Also, using Maclaurin expansion does not help here because we are in a neighborhood of $\infty$.
In addition, how can we prove that $\lim_{x\to\infty }(e^x-1-x-\frac{x^2}{2}) = \infty$ to start with (without using the $\varepsilon-\delta$ definition )?
Thank you!
Best Answer
Indeed, $\lim_{x\to\infty}1-\cos(x)$ doesn't exist. Nevertheless, and since $\lim_{x\to\infty}e^x-1-x=\infty$,$$\lim_{x\to\infty}\frac{1-\cos(x)}{e^x-1-x}=0,$$and therefore, by L'Hopital's rule, your limit is $0$ too.