How to find $\lim_{x\to\ln 2} \frac{2e^{3x}-16}{3e^{2x}-12}$? Need to find without using L'Hopital's rule or derivatives.
I know the answer is 2, but how can I find this analytically, without using limit tables?
Thanks!
 A: $$\lim_{x\to\ln 2} \frac{2e^{3x}-16}{3e^{2x}-12}$$
Take out some numbers:
$$\lim_{x\to\ln 2} \frac{2(e^{3x}-8)}{3(e^{2x}-4)}$$
Factor top and bottom (Numerator is difference of cubes, denominator is difference of squares):
$$\lim_{x\to\ln 2} \frac{2(e^x-2)(2e^x+e^{2x}+4)}{3(e^x-2)(e^x+2)}$$
You can cross out the $e^x-2$ terms and you are left with:
$$\lim_{x\to\ln 2} \frac{2(2e^x+e^{2x}+4)}{3(e^x+2)}$$
Now just plug it in:
$$\frac{2(2e^{\ln 2}+e^{2\ln 2}+4)}{3(e^{\ln 2}+2)}=\frac{2(2(2)+(4)+4)}{3((2)+2)}=\frac{24}{12}=\therefore 2$$
A: Let $u = e^x$. Then
\begin{align}
\lim_{x \rightarrow \ln 2} \frac{2e^{3x}-16}{3e^{2x}-12} &= \lim_{u \rightarrow 2} \frac{2u^3-16}{3u^2-12} &\text{substitute } u = e^{2x} \\
&= \lim_{u \rightarrow 2} \frac{2(u^3-8)}{3(u^2-4)} & \text{factor a 2 from top and a 3 from bottom} \\
&= \lim_{u \rightarrow 2} \frac{2(u-2)(u^2+2u+4)}{3(u+2)(u-2)} & \text{numerator: }(a^3-b^3)=(a-b)(a^2+ab+b^2)\\
&= \lim_{u \rightarrow 2} \frac{2(u^2+2u+4)}{3(u+2)} & (u-2)\text{'s cancel} \\
&= \frac{2((2)^2+2(2)+4)}{3((2)+2)} & \text{plug in $u=2$} \\
&= \frac{24}{12} = 2 \\
\end{align}
