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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.

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

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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 $x \to 0$.

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Hint: Split $\tan x$ as $\frac{\sin x}{\cos x}$ and try to break your limit into product of two finite limits

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    $\begingroup$ Thanks! Will try this before peeking at other answers :) $\endgroup$ – TheAlPaca02 Nov 13 '16 at 18:50
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You have

$$\tan(X)\sim X \;\;\;(X\to 0)\implies$$

$$\tan^2(\pi x)\sim (\pi x)^2\;\; (x\to 0)$$

$$\implies \lim_{x\to 0}\frac{\tan^2(\pi x)}{2(\pi x)^2}=\frac 12.$$

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  • $\begingroup$ True, but proving these things is non-trivial without derivatives $\endgroup$ – Brevan Ellefsen Nov 13 '16 at 21:39

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