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 would you use L'Hopital's Rule to find the limit of the following function ???

$$\lim_{n \rightarrow 0} \frac{1}{\pi x} \frac{\sin(\frac{\pi x}{2})} { e^{2\pi x}}$$

How would you define your $f(x)$ and $g(x)$ ?

Why cant I define $f(x) =\frac{ \sin(\frac{\pi x}{2})} {\pi x}$ and $g(x) = e^{2\pi x}$ instead??

share|cite|improve this question

If you want L'Hopital:

Let $f(x) = \sin\left(\frac{\pi x}{2}\right)$, $g(x) = \pi x e^{2\pi x}$. Thus

$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{\frac{\pi}{2} \cos\left(\frac{\pi x}{2}\right)}{\pi e^{2\pi x} (2\pi x +1)} = \frac{\pi}{2\pi} \cdot \frac{\cos 0}{e^0 (0+1)} = \frac{1}{2}$$

share|cite|improve this answer

I would say that the $e^{2\pi x}$ approaches $1$, so is harmless. Then you can deal with the rest.

So I would let $f(x)=\sin\left(\frac{\pi x}{2}\right)$ and $g(x)=\pi x$. We could equally well say that $\pi e^{2\pi x}$ has limit $\pi$, and let $f(x)$ be as before, and $g(x)=x$.

Remark: We can also keep things exactly as they are given to us, let $f(x)$ be the top, $g(x)$ the bottom. It is more work, for the differentiation of the bottom is more complicated, with a greater chance of error. And anyway sooner or later we will have to use the fact that $e^{2\pi x}$ approaches $1$.

share|cite|improve this answer

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.