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How does one take the limit of expressions involving logarithms?

$\displaystyle\lim_{x\rightarrow\infty} \ln(x)=$ ?

I know this diverges to infinity, but what if I was taking the natural log of something a bit more complicated than just $x$?

I was thinking that the limit as $x\rightarrow\infty$ of $\ln(e(x))$, where $e(x) = (1+1/x)^x$, surely should be $1$... does that mean that I can take the limit of whatever I am taking the $\log$ of, then take the $\log$ of that limit?

jimjim
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Daniq
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2 Answers2

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Since the logarithm is continuous, it's fine to say

$$\lim_{x\to\infty}\log(f(x))=\log\left(\lim_{x\to\infty}f(x)\right)$$

So

$$\lim_{x\to\infty}\log\left(\left(1+\frac1x\right)^x\right)=\log\left(\lim_{x\to\infty}\left(1+\frac1x\right)^x\right)=\log(e)=1$$

Alice Ryhl
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The case of $\log(e(x))$ can be dealt with in the following way: $$ \lim_{x\to\infty}\log\left(1+\frac{1}{x}\right)^{\!x}= \lim_{x\to\infty}x\log\left(1+\frac{1}{x}\right)= \lim_{t\to0^+}\frac{\log(1+t)}{t} $$ which is precisely the derivative of $x\mapsto\log(1+x)$ at $0$, because this is $$ \lim_{h\to0}\frac{\log(1+(0+h))-\log(1+0)}{h} $$ It depends now on what you're allowed to use. But if you set $\log(1+t)=u$, then $1+t=e^u$, so $t=e^u-1$ and the limit becomes $$ \lim_{u\to0^+}\frac{u}{e^u-1} $$ which is generally considered a known limit.

egreg
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