calculus - Solving a trig limit without L'Hopital: $lim_{x to 0} frac{tan^2 (pi x)}{2(pi x)^2}$

I'm supposed to solve this limit without using L'Hopitals rule.

I find the indeterminate form of $\frac{0}{0}$ which tells me that L'Hopitals is an option but since we haven't seen derivatives yet I'm not allowed to used it.

Previously I already tried substituting $\pi x$ for $t$ and dividing numerator and denominator by $\pi x$ both without success.

$$\lim_{x \to 0} \frac{\tan^2 (\pi x)}{2(\pi x)^2}$$

Answer

If one knows that
$$
\lim_{x\to 0} \frac{\sin x}{x}=1
$$ then one may write, as $x \to 0$,
$$
\frac{\tan^2(\pi x)}{2\pi^2 x^2}=\frac1{2}\cdot\left(\frac{\sin (\pi x)}{\pi x}\right)^2\cdot \frac1{\cos^2 (\pi x)}

$$ and conclude easily since $\displaystyle \frac1{\cos^2 (\pi x)} \to 1$ as $x \to 0$.