Finding the limit of $ \frac{\tan^2x-x^2}{x^2\tan^2x} $ How to find the limit of $ \frac{\tan^2x - x^2}{x^2\tan^2x}$ as $x$ approaches $0$.

I don't know how to approach and starting differentiating the numerator and denominator many times which after the 4th time gave me the result $\frac{2}{3}$. I want to know if there is any good simplification which can lead me to a simple and quick solution. Please give some ideas. 

Thank you.
Note that as $x \to 0$, we have $\tan(x) = x + \dfrac{x^3}{3} + O(x^5)$ (this is really a 2 line calculation). So,
\begin{align}
\tan^2(x) &amp;= x^2 + \dfrac{2x^4}{3} + O(x^6)
\end{align}
Hence,
\begin{align}
\dfrac{\tan^2x - x^2}{x^2 \tan^2x} &amp;= \dfrac{\frac{2x^4}{3} + O(x^6)}{x^4 + O(x^6)} \\
&amp;= \dfrac{\frac{2}{3} + O(x^2)}{1 + O(x^2)},
\end{align}
where I divided top and bottom by $x^4$. Thus, if we take the limit $x \to 0$, the answer is $\dfrac{2}{3}$.



By the way, observe that it is no coincidence you had to differentiate $4$ times, because as you can see in the above derivation, the Taylor polynomials of the numerator and denominator begin at the $x^4$ term, so it is precisely the ratio of these coefficients which you're calculating if you use something like L'Hopital's rule.
Recall that $\lim_{x\to 0}\dfrac{\sin(x)}{x}=1$. The proof of this involves the definition of trigonometric functions by unit circle, which I will omit here. An immediate consequence of this is $\lim_{x\to 0}\dfrac{\tan(x)}{x}=1$. Hence dividing both the numerator and the denominator by $x^2$, we see that the desired limit is indeed 0.