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Solve $$\lim\limits_{x\to0}\left(1+\frac{1}{2x}\right)^x$$

i tried solve the limit by this way

$$\begin{align} \lim\limits_{x\to0}\left(1+\frac{1}{2x}\right)^x&\stackrel{?}{=}\left[\lim\limits_{x\to0}\left(1+\frac{1}{2x}\right)\right]^{\lim\limits_{x\to0}x}\\ &=(\pm\infty)^0\\ &\stackrel{?}{=}1 \end{align}$$

but im not sure if i made it correct, how i solve it?

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    $\begingroup$ If both limits were finite, that might work, but this won't work when the value is undefined. $\endgroup$ Commented Apr 7, 2015 at 21:30

2 Answers 2

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Certainly, the term is bigger than 0 in some neighborhood of 0, and thus we can safely take the logarithm.

$L'=\lim_{x\to0}x\ln{(1+\frac{1}{2x})}=\lim_{x\to0}\frac{\ln(1+\frac{1}{2x})}{\frac{1}{x}}$

Now, we use L'hopital's rule:

$L'=\lim_{x\to0}\frac{\frac{-2}{x^2+2x}}{\frac{-1}{x^2}}=\lim_{x\to0}\frac{2x^2}{x^2+2x}=0$

Now, exponentiating, we get

$L=e^{L'}=1$

QED.

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Set $\frac{1}{x}=n$ and rewrite the whole expression as $e^{\log L}$. What do you get?

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    $\begingroup$ That's a risking substitution. since then you need to compute $n\to \pm\infty$. $\endgroup$ Commented Apr 7, 2015 at 21:32
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    $\begingroup$ if i start with logging, only positive values are possible $\endgroup$
    – Alex
    Commented Apr 7, 2015 at 21:37
  • $\begingroup$ @Alex no, $n$ is outside the log. $\endgroup$ Commented Apr 7, 2015 at 22:16

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