Jasmin Pineda

2022-06-19

Solving a trig limit without L'Hopital: $\underset{x\to 0}{lim}\frac{{\mathrm{tan}}^{2}(\pi x)}{2(\pi x{)}^{2}}$

Aiden Norman

Beginner2022-06-20Added 16 answers

If one knows that

$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$

then one may write, as $x\to 0$,

$\frac{{\mathrm{tan}}^{2}(\pi x)}{2{\pi}^{2}{x}^{2}}=\frac{1}{2}\cdot {\left(\frac{\mathrm{sin}(\pi x)}{\pi x}\right)}^{2}\cdot \frac{1}{{\mathrm{cos}}^{2}(\pi x)}$

and conclude easily since $\frac{1}{{\mathrm{cos}}^{2}(\pi x)}\to 1$ as $x\to 0$

$\underset{x\to 0}{lim}\frac{\mathrm{sin}x}{x}=1$

then one may write, as $x\to 0$,

$\frac{{\mathrm{tan}}^{2}(\pi x)}{2{\pi}^{2}{x}^{2}}=\frac{1}{2}\cdot {\left(\frac{\mathrm{sin}(\pi x)}{\pi x}\right)}^{2}\cdot \frac{1}{{\mathrm{cos}}^{2}(\pi x)}$

and conclude easily since $\frac{1}{{\mathrm{cos}}^{2}(\pi x)}\to 1$ as $x\to 0$

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