Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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...

share|cite|improve this question
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
up vote 5 down vote accepted

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....

share|cite|improve this answer
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.

share|cite|improve this answer
But where was my error? – user1009013 Nov 7 '11 at 15:32
1's a well-established limit, pedja. – J. M. 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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.