How can I calculate the limit $\lim_{x\to-2}\frac{\tan (\pi \cdot x)}{(x+2)}$ without l'Hopital?

Question

How can I calculate $\frac{\tan (\pi \cdot x)}{(x+2)}$ as $x \to -2$ without the rule of L'Hopital? When I try, I get infinity... But the correct answer is $\pi$
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I split the tan into sin/cos and multiply and divide by $2 \cos(\pi \cdot x)$, so I get $\cos (\pi \cdot x \cdot 2)$ above and $(2 \cos( \pi \cdot x)^2) \cdot (x+2)$ below. So I become 1/0 and thus infinity...

Answer 1

You can also observe that your limit is

$$\lim_{x \to -2} \frac{\tan (\pi \cdot x)}{(x+2)\pi} \pi =\lim_{x \to -2} \frac{\tan (\pi \cdot x)- \tan(-2 \pi)}{(x\pi - (-2)\pi)} \pi \,.$$

Denoting $y=\pi \cdot x$ you have

$$\lim_{x \to -2} \frac{\tan (\pi \cdot x)- \tan(-2 \pi))}{(x\pi - (-2)\pi}=\lim_{y \to -2 \pi} \frac{\tan (y)- \tan(-2 \pi)}{(y - (-2)\pi}$$

which is just the definition of the derivative....

Answer 2

Write it as $$\frac{\tan(\pi x)}{x+2} = \frac{\sin(\pi x)}{\cos(\pi x)(x+2)} = \frac{\sin(\pi(x + 2))}{\cos(\pi x)(x+2)} = \frac{\pi \sin(\pi(x+2))}{\cos(\pi x)\pi(x+2)}.$$ Now as $x \to -2$, $x+2 \to 0$ and $\sin(\pi(x+2))/(\pi(x+2)) \to 1$ and $\cos(\pi x) \to 1$, so we get the result.