Calculate $\lim_{x \rightarrow 0} \frac{e^{x}-x-1}{1-\cos x}$

Question

I know how to do this task and my answer is $1$.

However I think that my way is a long way to the goal and I want to try do this task using Taylor's theorem with the rest in the form of Peano.

Unfortunately I do not fully understand how I can do it.

Can you help me?

Answer 1

Using

• $e^x = 1+x+\frac{x^2}{2} + o(x^2)$ and
• $\cos x = 1 - \frac{x^2}{2} + o(x^2)$

you get $$\frac{e^{x}-x-1}{1-\cos x} = \frac{\frac{x^2}{2} + o(x^2)}{\frac{x^2}{2} + o(x^2)}\stackrel{x \to 0}{\longrightarrow}1$$

Answer 2

Using expansions up to order $2$: $$\lim_{x\to 0} \frac{e^x-x-1}{1-\cos x} = \lim_{x \to 0}\frac{1+x+x^2/2+o(x^2)-x-1}{1-1+x^2/2+o(x^2)} = \lim_{x \to 0} \frac{1/2+o(x^2)/x^2}{1/2+o(x^2)/x^2} = 1. $$

Answer 3

The simpest way is to apply L'Hopital's Rule twice.

Answer 4

Using l'Hospital: $$\lim_{x\rightarrow 0}\frac{\exp(x)-x-1}{1-\cos(x)}=\lim_{x\rightarrow 0}\frac{\exp(x)-1}{\sin(x)}=\lim_{x\rightarrow 0}\frac{\exp(x)}{\cos(x)}=\frac{\exp(0)}{\cos(0)}=1.$$

Answer 5

$$\lim_{x\to0}\dfrac{e^x-1-x}{1-\cos x}=\lim_{x\to0}\dfrac{e^x-1-x}{x^2}\lim_{x\to0}\left(\dfrac x{\sin x}\right)^2\lim_{x\to0}(1+\cos x)$$

For the first limit use Are all limits solvable without L'Hôpital Rule or Series Expansion