I'm supposed to solve this limit without using L'Hopitals rule.
I find the indeterminate form of 00 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 πx for t and dividing numerator and denominator by πx both without success.
limx→0tan2(πx)2(πx)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.
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