Limits of power of fractions I'm having trouble solving these limits:
$$\lim_{x\to \infty} \left (\frac{3-x}{2x+5}\right )^{3x}\\\lim_{x\to \infty} \left (\frac{2x+1}{3x-4}\right )^{1-2x}\\\lim_{x\to \infty} \left (\frac{2-3x}{1-3x}\right )^{x+4}$$
Are they even defined?
 A: Hints:
Good question: the fractions have to be positive when $|x|$ gets larger and larger. Now it's easy the check the first fraction, $\dfrac{3-x}{2x+5},\,$ is positive if and only if $\;-\frac52<x<3$, so the first limit doesn't make sense.
For the other two, it makes sense (at $+\infty$ and $-\infty$). You shoould determine the limits of the logs, after you have written the fractions as
\begin{align}
\frac{2x+1}{3x-4}&=\frac23\,\frac{x+\frac12}{x-\frac43}=\frac23\biggl(1+\frac {11}{2(3x-4)}\biggr),&\qquad
\frac{x-\frac23}{x-\frac13}&= 1-\frac1{3x-1}.
\end{align}
Finally you'll have to use  $\;\log(1+u)=1+u+o(u)$ near $u=0\;$ (or $\log(1+u)\sim_0u$).
A: Hint: Knowing that $$e^a =\lim_{x\to \infty}\left(1+\frac{a}{x}\right)^x$$ 
show by yourself that: 
$$\lim_{x\to \infty} \left (\frac{3-x}{2x+5}\right )^{3x}~~~~~~~~DNE$$ 
$$\lim_{x\to \infty} \left (\frac{2x+1}{3x-4}\right )^{1-2x}= \lim_{x\to \infty} \left (\frac{3}{2}-\frac{11}{4x+2}\right )^{2x-1}=\infty$$ 
$$\lim_{x\to \infty} \left (\frac{2-3x}{1-3x}\right )^{x+4}=\lim_{x\to \infty} \left (1-\frac{1}{3x-1}\right )^{3x-1 \frac{x+4}{3x-1}} =e^{-\frac{1}{3}}$$
use similar idea as here Proving that $\lim_{n\to\infty}\left ( \frac{2n-1}{2n+3} \right )^n=e^{-2}$ without using de l‘Hôspital
