Calculating $\lim_{x\to0}\frac{e^{\cos x}-e}{x^2}$ without L'Hôpital

Question

Calculate$$\lim_{x\to0}\frac{e^{\cos x}-e}{x^2}$$without L'Hôpital.

I tried changing $e$ to $(1+x)^{1/x}$, tried using $\frac{1-\cos x}{x^2} = \frac{1}{2}$, tried converting $\cos x$ to $1-2\left(\sin \frac{x}{2}\right)^2$ but I always got stuck at some point, and would really appreciate help.

Answer 1

Hint:

For $x(\cos x-1)\ne0,$ $$\dfrac{e^{\cos x}-e}{x^2}=-e\cdot\dfrac{e^{(\cos x-1)}-1}{\cos x-1}\cdot\dfrac{1-\cos x}{x^2}$$

Now $\lim_{h\to0}\dfrac{e^h-1}h=?$

Answer 2

Not as elegant as @lab bhattacharjee's solution, but still a pretty short solution. Set

$$L=\lim_{x\to0}\frac{e^{\cos x}-e}{x^2}=e\cdot\lim_{x\to0}\frac{e^{\cos x-1}-1}{x^2}.$$

Now using that

$$e^t=1+t+\mathcal{O}(t^2)$$

and

$$\cos t=1-\frac{t^2}{2}+\mathcal{O}(t^4)$$

as $t\to0$, we have that

$$L=e\cdot\lim_{x\to0}\frac{-\frac{x^2}{2}+\mathcal{O}(x^4)}{x^2}=-\frac{e}{2}.$$

Answer 3

Let $f(t)=e^{\cos\sqrt t}.$ $$\lim_{t\to0^+}f'(t)=\lim_{t\to0^+}\left(-\frac{f(t)}2\frac{\sin\sqrt t}{\sqrt t}\right)=-\frac{f(0)}2\cdot1=-\frac e2$$ hence $$\lim_{x\to0}\frac{e^{\cos x}-e}{x^2}=\lim_{x\to0^+}\frac{e^{\cos x}-e}{x^2}=\lim_{t\to0^+}\frac{f(t)-f(0)}t=f'(0)=-\frac e2.$$