<s>[INST] 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.
 [/INST] 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 $-\frac 12\leq x \leq \frac 12$.
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