# Evaluate the limit using power series without L'Hospital's Rule [closed]

I'm a bit stumped on this one.

Show that $$\lim_{x\to0} \frac{e^x -1}{\sin(x)} = 1$$ using power series.

The instructions are not to use L'Hospital's Rule. I cannot find a way to do this without L'Hopital even simplifying using series expansion.

## closed as off-topic by RRL, Mark Viola, Leucippus, José Carlos Santos, ancientmathematicianApr 24 at 7:03

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$$\lim_{x\to0} \frac{e^x -1}{\sin(x)} =\lim_{x\to0} \frac{-1+1+x+\frac{x^2}{2!}+...}{x-\frac{x^3}{3!}+...}=\lim_{x\to0}\frac{x(1+\frac{x}{2!}+...}{x(1-\frac{x^2}{3!}+...} = 1$$
A simpler solution in my opinion using fundamental limits and the definition of the derivative. Note that $$\lim_{x\to 0}\frac{e^x-1}{\sin x}= \lim_{x\to 0}\left( \frac{e^x-1}{x}\times \frac{x}{\sin x} \right)=\lim_{x\to 0}\frac{e^x-1}{x}\times\lim_{x\to 0}\frac{x}{\sin x} =\frac{d}{dx}(e^x)\bigg|_{x=0}\times 1=1$$
$$\frac{\mathrm e^x-1}{\sin x}=\underbrace{\frac{\mathrm e^x-1}{x}}_{\substack{\downarrow\\\tfrac{\mathrm{d(e}^x)}{\mathrm dx}\Bigm\vert_{x=0}=1}} \underbrace{\frac x{\sin x}}_{\substack{\downarrow\\1}}$$