# How can I calculate a Limit without l'Hopital?

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 $\pi$
:(

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...

-
Why not use the rule of l'Hopital ? Anyway, in your calculation, I think you should have $\cos(\pi x) \sin(\pi x) = \frac{1}{2}\sin(2\pi x)$ above. This yields the indeterminate form $\frac{0}{0}$ once again. –  Joel Cohen Nov 7 '11 at 15:31

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....

-
Accepted within 3 minutes... a talking point over in chat :) –  The Chaz 2.0 Nov 7 '11 at 15:38
Personally I think the other solution is better ;) –  N. S. Nov 7 '11 at 15:42
(Mine is just a commentary on the practice of immediately accepting answers. I personally would have gone the route of the other answer, which is why I appreciate yours!) –  The Chaz 2.0 Nov 7 '11 at 15:47
@TheChaz Yea... We probably won't see him again until the next question... And the funny part is that there could had been a mistake in my solution, I saw in the past few wrong answers upvoted, because the mistake was subtle. –  N. S. Nov 7 '11 at 15:51

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.

-
But where was my error? –  user1009013 Nov 7 '11 at 15:32
...it's a well-established limit, pedja. –  J. M. is back. Nov 7 '11 at 15:33
Isn't the limit of $\frac{\sin x}{x}$ obtained using l'Hopital's rule again ? –  Joel Cohen Nov 7 '11 at 15:34
@user1009013: You should have got $\sin(2\pi x)$ in the numerator. –  J. J. Nov 7 '11 at 15:34
Btw, the exercise was to solve it without l'hopital... in other exercises, I almost always use l'hopital, it's so powerful... –  user1009013 Nov 7 '11 at 15:34