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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?

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closed as off-topic by Guy Fsone, Namaste, Cameron Williams, Claude Leibovici, Shailesh Jan 25 '18 at 9:43

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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

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  • $\begingroup$ Thanks! Managed to solve two of them, I got one more question however. Applying the same rule on the first one leads me to $$ \left (e^{-3x-2} \right )^{\frac{3}{2}} = e ^{-\infty} = 0$$ . Where do I get it wrong? $\endgroup$ – Atti Jan 25 '18 at 1:05
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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 $\log(1+u)\sim_0u$).

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