I would like to evaluate the following limit: $$\lim_{x \to 0}\frac{\tan^2x-x^2}{x^2\tan^2x}$$
I tried breaking the limit as:
$$\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{1}{\tan^2x}\right)$$
Writing tan is terms of sin and cos, this becomes:
$$\lim_{x \to 0}\left(\frac{1}{x^2}-\frac{\cos^2x}{\sin^2x×x^2/x^2}\right)$$
Using standard limit of $\sin x/x$, we obtain:
$$\lim_{x \to 0}\frac{1-\cos^2x}{x^2}$$
Since $1-\cos^2x=\sin^2x$, using standard limits again we get 1 as the answer. But the correct answer is $2/3$. Please explain what is going wrong.