How to solve this limit without L'Hospital's rules

Question

I'd needed to solve limit without the said rules:

$\lim_{x\to0}\dfrac{1-e^{x^2}}{1-\cos(x)}$

I wonder if all limits can be done without applying L'Hospital's rules (not Taylor series either). Could anyone help me some hints to solve this? Thanks in advance.

Answer 1

First:

Rewrite your limit as: $\lim_{x\rightarrow0}\frac{\displaystyle-\frac{-1+e^{x^2}}{x^2}x^2}{\displaystyle\frac{1-\cos(x)}{x^2}x^2}$ and using remarkable limits this tends to $-2$.

As for your other question: Yes. All the limits can be done without using L'hospital. Let's assume that there exists a limit that can be done only by using L'hospital rule but that limit can be proven by proving the L'hospital rule therefore such a limit does not exist. In conclusion you can solve any limit without using L'hospital rule, but sometimes it's indicated to use L'hospital rule because it's easier to solve the question and you always want to shortest way to the answer.

Answer 2

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

Answer 3

HINT

Recall the standard limits as $t\to 0$

• $\frac{e^t-1}{t}\to 1$
• $\frac{1-\cos t}{t^2}\to \frac12$

Answer 4

\begin{align*} \dfrac{1-e^{x^{2}}}{1-\cos x}&=\dfrac{1-e^{x^{2}}}{2\sin^{2}(x/2)}\\ &=-\dfrac{e^{x^{2}}-1}{x^{2}}\cdot\left(\dfrac{x/2}{\sin(x/2)}\right)^{2}\cdot 2, \end{align*} now \begin{align*} e^{x^{2}}-1=\dfrac{1}{x^{2}}\int_{0}^{x^{2}}e^{u}du=e^{\eta_{x}}\rightarrow 1 \end{align*} as $x\rightarrow 0$, the limit is $-2$.