Compute $\lim_{x\rightarrow0}\frac{e^{x^2} – \cos x}{\sin^2 x}$

Find the limit as $x$ approaches $0$ of

$\dfrac{e^{x^2} – \cos x}{\sin^2 x}$

What I tried is as $x$ approaches $0$, $e^{x^2}$ tends to $1$ and so the numerator tends to $1-\cos x$ and after doing some trigonometric simplifications I got the answer as $\frac 12$.

Is this right? Any help would be appreciated.

Solutions Collecting From Web of "Compute $\lim_{x\rightarrow0}\frac{e^{x^2} – \cos x}{\sin^2 x}$"

Using l’Hospital:

$$\lim_{x\to 0}\frac{e^{x^2}-\cos x}{\sin^2 x}\stackrel{\text{l’H}}=\lim_{x\to 0}\frac{2xe^{x^2}+\sin x}{2\sin x\cos x}\stackrel{\text{l’H}}=\lim_{x\to 0}\frac{2e^{x^2}+4x^2e^{x^2}+\cos x}{2\cos^2x-2\sin^2x}=\frac{2+0+1}{2}=\frac32$$

HINT:

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

Use $\lim_{h\to0}\dfrac{e^h-1}h=1=\lim_{u\to0}\dfrac{\sin u}u$ for the first part

and as $x\to0,\cos x\to1\implies\cos x-1\ne0$ for the second

Use Taylor expansion. As $x\rightarrow0$, you have

$$e^{x^2}=1+x^2+O(x^4)$$
$$\cos x=1-\frac12x^2+O(x^4)$$
$$\sin^2x=x^2+O(x^4)$$

Thus

$$\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{1+x^2-1+\frac12x^2+O(x^4)}{x^2+O(x^4)}=\frac{\frac32+O(x^2)}{1+O(x^2)}\underset{x\rightarrow 0}{\longrightarrow} \frac32$$

As Jean-Claude Arbaut showed, Taylor expansions are extremely useful for this kind of problems.

But, using them with a few more terms, you also see how is the limit approached $$A=\frac{e^{x^2}-\cos x}{\sin^2 x}=\frac{\Big(1+x^2+\frac{x^4}{2}+O\left(x^6\right)\Big)-\Big(1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)\Big)}{x^2-\frac{x^4}{3}+O\left(x^6\right)}$$ $$A\approx\frac{\frac{3 x^2}{2}+\frac{11 x^4}{24}+O\left(x^6\right)}{x^2-\frac{x^4}{3}+O\left(x^6\right)}=\frac{\frac{3 }{2}+\frac{11 x^2}{24}}{1-\frac{x^2}{3}}$$ Performing the long division, we have $$A\approx \frac{3}{2}+\frac{23 }{24}x^2$$ If you plot on the same graph the function and the above approximation, you will probably be amazed to see how close to each other are the two curves for $-\frac 12\leq x \leq \frac 12$.

You can use Taylor series as

$$\dfrac{e^{x^2} – \cos x}{\sin^2 x} \sim_{x\sim 0} \dfrac{(1+x^2) – (1-x^2/2)}{x^2} =\frac{3}{2}. $$