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3ds max 2013 serial number and product key free – Huge customer supportQ:

Show that $\lim_{x\to 0}\frac{e^{x^2}-1}{x}=1$?

Show that $\lim_{x\to 0}\frac{e^{x^2}-1}{x}=1$. I tried many methods and I found none that worked. Please, I would be grateful if anyone helps me.

A:

The limit can be easily found to be
$$\lim_{x\to0} \frac{e^{x^2}-1}{x} = \lim_{x\to0} \frac{\left(e^{x}-1\right)\left(e^{ -x}-1\right)}{x\cdot e^{ -x}}
= \lim_{x\to0} \frac{e^{ -x}-1}{ -x}$$
so it’s enough to find a limit of $1/(-x)$. A standard trick works here, thanks to the fact that
$$\exp(x)-1 = \exp(x)\cdot\frac{1}{\exp(x)-1}$$
We have
$$\frac{1}{x} = \lim_{x\to0} \frac{ -1}{x\cdot\exp(x)}
= \lim_{x\to0} \frac{\left(\exp(x)-1\right)}{ -x}$$
so
$$\lim_{x\to0} \frac{1}{ -x} = \lim_{x\to0} \frac{\left(\exp(x)-1\right)}{ -x\cdot\left(\exp(x)-1\right)} = \lim_{x\to0} \frac{1}{ -x\cdot\left(\exp(x)-1\right)} = \lim_{x\to0} \frac{1}{\exp(x)-1}$$

A:

Without any limits, we know that
$$\frac{e^{x^2}-1}{x}=\frac{e^{2x}-1}{2x}=e^{x}\cdot\frac{e^{ -x}-1}{ -x}$$
Thus
$$\lim
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