A limit problem where $x\to 0$ I can't figure out how to compute this limit: $$\lim_{x\to 0}\dfrac{e^{3x^2}-e^{-3x^2}}{e^{x^2}-e^{-x^2}}$$ Any advice?
 A: Setting $t=e^{x^2}$ gives
$$\lim_{x\to 0}\frac{e^{3x^2}-e^{-3x^2}}{e^{x^2}-e^{-x^2}}=\lim_{t\to 1}\frac{t^3-\frac{1}{t^3}}{t-\frac 1t}=\lim_{t\to 1}\frac{t^6-1}{t^4-t^2}=\lim_{t\to 1}\frac{(t-1)(t^2+t+1)(t^3+1)}{t^2(t-1)(t+1)}$$
A: Notice that 
$$e^{3x^{2}}-e^{-3x^{2}} = (e^{x^{2}})^{3} - (e^{-x^{2}})^{3} = (e^{x^{2}}-e^{-x^{2}})(e^{2x^{2}}+e^{-2x^{2}} +1)$$ 
Now use this to cancel out potential problems in the denominator
A: Notice, $$\lim_{x\to 0}\frac{e^{3x^2}-e^{-3x^2}}{e^{x^2}-e^{-x^2}}$$
$$\lim_{x\to 0}\frac{e^{3x^2}-\frac{1}{e^{3x^2}}}{e^{x^2}-\frac{1}{e^{x^2}}}$$
$$=\lim_{x\to 0}\frac{e^{x^2}(e^{6x^2}-1)}{e^{3x^2}(e^{2x^2}-1)}$$
$$=\lim_{x\to 0}e^{-2x^2}\times 3\lim_{x\to 0}\frac{\frac{(e^{6x^2}-1)}{6x^2}}{\frac{(e^{2x^2}-1)}{2x^2}}$$
$$=3\lim_{x\to 0}e^{-2x^2}\times \frac{\lim_{x\to 0}\frac{e^{6x^2}-1}{6x^2}}{\lim_{x\to 0}\frac{e^{2x^2}-1}{2x^2}}$$
$$=3(1)\times \frac{1}{1}=\color{red}{3}$$
A: There are four ways:


*

*Use L'Hospital's rule directly

*Use Taylor series for $\exp(x)=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$, this is in most cases the best way

*This is the most basic way to calculate the limit: Use 
$$\lim_{x\to 0}\frac{e^{3x^2}-e^{-3x^2}}{e^{x^2}-e^{-x^2}}=\lim_{x\to 0}\frac{(e^{x^2})^3-(e^{-x^2})^3}{e^{x^2}-e^{-x^2}}$$
$$=\lim_{x\to 0}\frac{(e^{x^2}-e^{-x^2})((e^{x^2})^2+e^{x^2}e^{-x^2}+(e^{-x^2})^2)}{e^{x^2}-e^{-x^2}}$$
$$=\lim_{x\to 0}[(e^{x^2})^2+e^{x^2}e^{-x^2}+(e^{-x^2})^2]=1+1+1=3$$

*Using $\sinh(x)=\frac{e^x-e^{-x}}{2}$ rewrite the fraction and use L'Hospital's rule
$$\lim_{x\to 0}\frac{\sinh(3x^2)}{\sinh(x^2)}=\lim_{x\to 0}\frac{6x\cosh(3x^2)}{2x\cosh(x^2)}=\frac{3\cdot1}{1}=3$$
A: There are many ways. For example we can use plain algebra. Multiply top and bottom by $e^{3x^2}$. We get
$$\frac{e^{6x^2}-1}{e^{2x^2}(e^{2x^2}-1)}.$$
Let $u=e^{2x^2}$. Then the top is $u^3-1$ and the bottom is $e^{2x^2}(u-1)$. Factor. We get $\frac{u^2+u+1}{e^{2x^2}}$ and the rest is easy.
Remark: Series are much more natural but more "advanced."
A: $$\lim_{x\to 0}\dfrac{e^{3x^2}-e^{-3x^2}}{e^{x^2}-e^{-x^2}}$$
$$=\lim_{x\to 0}\dfrac{e^{3x^2}-1-\left(e^{-3x^2}-1\right)}{e^{x^2}-1-\left(e^{-x^2}-1\right)}$$
$$=\dfrac{3\cdot\lim_{x\to 0}\dfrac{e^{3x^2}-1}{3x^2}-\left(\lim_{x\to 0}\dfrac{e^{-3x^2}-1}{-3x^2}\right)(-3)}{\lim_{x\to 0}\dfrac{e^{x^2}-1}{x^2}-\left(\lim_{x\to 0}\dfrac{e^{-x^2}-1}{-x^2}\right)(-1)}$$
Now use $\lim_{h\to0}\dfrac{e^h-1}h=1$
