# limit of $\frac{e^{2x}-x^2+x}{\cos(x)-1}$ as $x\to 0$.

The title said it, what is: $$\lim_{x\to 0} \frac{e^{2x}-x^2+x}{\cos(x)-1} = ~?$$ If I evaluate the term I get $1/0$, by looking at a graph I see that it goes to $-\infty$, but I don't now how evaluate the limit?

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The numerator tends to 1.

I would approximate the denominator, $\cos(x) \approx 1 - x^2/2$ and $\cos x - 1 \approx - x^2/2 < 0$

In limit, we can make these substitutions: $\lim_{x\to 0} \frac{e^{2x}-x^2+x}{\cos(x)-1} = \lim_{x\to 0} \frac{-2}{x^2} = - \infty$

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The limit of the numerator is $1$.

The limit of the denominator is $0^-$.

So the limit of your function is $-\infty$.

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+1 for this tiny approach. ;-) – Babak S. Feb 1 '13 at 6:01
what you mean by limit is $0^{-}$? i think the limit is just $0$ in the usual definition of limit... – Stefan Feb 1 '13 at 19:58
I mean the limit is $0$, with negative values close to $0$ (and for all $x\neq 0$ actually here). It approaches $0$ from the left, if you prefer. Like when you distinguish the left limit and the right limit at some point. – 1015 Feb 1 '13 at 20:03

Hint: Note that when $\alpha(x)$ is very small while $x\to\infty$ then $$\text{e}^{\alpha(x)}-1\sim\alpha(x), ~~1-\cos(\alpha(x))\sim(\alpha^2(x))/2$$ Of course, $\lim_{x\to 0}f(x)=-\infty$.

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Nice approach for developing intuition about problems like this! + – amWhy Feb 1 '13 at 1:20