<s>[INST] What did I do wrong trying to find this limit? In another question, a user asked to find:
$$\lim_{x\to 0} \frac{\exp(x^2)-\cos(x)}{\sin(x)^2}$$
I thought I could use pure trigonometric identities to find the limit. Apparently I was mistaken, but I can't find my mistake. Here's what I did:
Knowing $\sin(x)^2 = \frac{1-\cos(2x)}{2}$:
$$\frac{\exp(x^2)-\cos(x)}{\sin(x)^2} = \frac{\exp(x^2)-\cos(x)}{\frac{1-\cos(2x)}{2}} = \frac{\exp(x^2)-\cos(x)}{1} \cdot \frac{2}{1-\cos(2x)}$$
$$ = \frac{2(\exp(x^2)-\cos(x))}{1-\cos(2x)}$$
And knowing $\cos(2x) = 2\cos(x)^2-1$:
$$\frac{2(\exp(x^2)-\cos(x))}{1-\cos(2x)} = \frac{2(\exp(x^2)-\cos(x))}{1-(2\cos(x)^2-1)} = \frac{2(\exp(x^2)-\cos(x))}{-2(\cos(x)^2)}$$
$$ = -1 \cdot \frac{\exp(x^2)-\cos(x)}{\cos(x)^2} = -1 \cdot \left( \frac{\exp(x^2)}{cos(x)^2} -\frac{\cos(x)}{\cos(x)^2} \right)
= -1 \cdot \left( \frac{\exp(x^2)}{cos(x)^2} -\frac{1}{cos(x)} \right) $$
And we can plug in $x=0$ since we don't run the risk of an indeterminate form:
$$ -1 \cdot \left( \frac{\exp(0^2)}{cos(0)^2} -\frac{1}{cos(0)} \right)  = -1 \cdot \left( \frac{1}{1^2} -\frac{1}{1} \right) = -1 \cdot 0 = 0$$
Apparently the correct answer is $3/2$. Where did I go wrong?
 [/INST] The wrong step is $1-(2\cos^2x-1)=2-2\cos^2x\neq -2\cos^2x$.
Notice that you are simply itroducing double angle, then removing it, so it's equivalent to use the identity $\cos^2x +\sin^2x=1$.
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