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!

$$\lim_{x\to\ln 2} \frac{2e^{3x}-16}{3e^{2x}-12}$$
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:
$$\frac{2(2e^{\ln 2}+e^{2\ln 2}+4)}{3(e^{\ln 2}+2)}=\frac{2(2(2)+(4)+4)}{3((2)+2)}=\frac{24}{12}=\therefore 2$$
