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Find $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$.

$\displaystyle \lim_{x\to -\infty}\left(\frac{\ln(2^x + 1)}{\ln(3^x+ 1)}\right)$ = $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \ln(1+\frac{1}{2^x})}{\ln3^x + \ln(1+\frac{1}{3^x})}\right)$ = $\displaystyle \lim_{x\to -\infty}\left(\frac{\ln2^x + \frac{1}{2^x}\ln(1+\frac{1}{2^x})^{2^{x}}}{\ln3^x + \frac{1}{3^x}\ln(1+\frac{1}{3^x})^{3^{x}}}\right)$ = $\cdots$

This method is not working.

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  • $\begingroup$ Hint: Make a graph. At least you get an idea that probably you wouldn't have to go into an algebraic forest $\endgroup$ – imranfat Nov 12 '16 at 20:54
  • $\begingroup$ When $x \to -\infty$, then $2^x \to 0$. Look at $\ln (1 + z)$ for $z$ near $0$. $\endgroup$ – Daniel Fischer Nov 12 '16 at 20:55
  • $\begingroup$ @DanielFischer But then i have $\lim_{x \to - \infty} \frac{0}{0}$ $\endgroup$ – Nowak Grzegorz Nov 12 '16 at 21:00
  • $\begingroup$ But you know (roughly) how fast the numerator and the denominator tend to $0$. $\endgroup$ – Daniel Fischer Nov 12 '16 at 21:01
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    $\begingroup$ math.stackexchange.com/questions/1446831/… $\endgroup$ – Workaholic Nov 13 '16 at 11:17
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If we use the well known limit

$$\lim_{X\to 0}\frac{\ln(1+X)}{X}=1$$

with $X=2^x\;$ and $\;X=3^x,\;\;$we find

$$\lim_{x\to-\infty}\left( \frac{2}{3} \right)^x=+\infty$$

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    $\begingroup$ Sharp answer +1 $\endgroup$ – imranfat Nov 12 '16 at 21:05
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    $\begingroup$ Hahaha, no American slang.. Your answer only took few lines, which is great! $\endgroup$ – imranfat Nov 12 '16 at 21:08
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We have $$ \ln(1+t)=t+o(t^2). $$ Thus, since $2^x\to0$ and $3^x\to0$ when $x\to-\infty$, $$ \frac{\ln(1+2^x)}{\ln(1+3^x)}=\frac{2^x+o(2^{2x})}{3^x+o(3^{2x})} =\left(\frac23\right)^x\,\frac{1+o(2^x)}{1+o(3^x)}. $$ The second fraction goes to $1$ when $x\to-\infty$. So the limit will be equal to $$ \lim_{x\to-\infty}\left(\frac23\right)^x=\lim_{x\to-\infty}\exp\left(x\ln\left(\frac23\right)\,\right)=+\infty. $$

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