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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??

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2 Answers 2

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 $1$.

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If you want L'Hopital:

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

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

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