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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!

Joshua Rectangle
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  • By definition, $$ e^x - 1 - x - \frac{{x^2 }}{2} = \frac{{x^3 }}{6} + \frac{{x^4 }}{{24}} + \cdots > \frac{{x^3 }}{6}. $$ – Gary Sep 17 '21 at 18:30
  • $\lim_{x \to \infty} \frac{e^x}{x^2} = \infty$ by L'Hopital, now use this result , along with product laws to get a proof for the second result. Of course L'Hopital doesn't apply there directly, so you will have to be a little roundabout regarding this. – Sarvesh Ravichandran Iyer Sep 17 '21 at 18:31

2 Answers2

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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.

José Carlos Santos
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  • Thanks @Jose Carlos! Can you help me also figure out how to use l'Hospital rule to prove that $\lim_{x\to\infty} (e^x-1-x) = \infty$? – Joshua Rectangle Sep 17 '21 at 18:24
  • You will find several answers here. – José Carlos Santos Sep 17 '21 at 18:27
  • @JoséCarlosSantos: I would suggest that you make it explicit that if $f$ and $g$ are differentiable near $a$, and $\lim_{x\to a}g(x)=\infty$ (or $-\infty$), then we can apply L'Hôpital's rule to a limit of the form $\lim_{x\to a}f(x)/g(x)$ (it need not be the case that $\lim_{x\to a}f(x)=\infty$; in fact, this limit might not even exist). – Joe Sep 17 '21 at 21:03
  • @JoshuaRectangle: See my above comment. To apply L'Hôpital's rule, it doesn't need to be the case that the numerator tends towards infinity. – Joe Sep 17 '21 at 21:04
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As an alternative

$$\frac{x-\sin(x)}{e^x-1-x-\frac{x^2}{2}}=\frac{x}{e^x}\frac{1-\frac{\sin(x)}x}{1-\frac 1{e^x}-\frac{x}{e^e}-\frac{x^2}{2e^x}} \to 0\cdot 1=0$$

and we can use squeeze theorem

$$\left|\frac{\sin(x)}x\right|\le \frac 1 x \to 0$$

and l'Hospital for each term to show that $\frac{x^n}{e^x} \to 0$ or also referring to

user
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