I'm trying to figure out how the $\displaystyle \lim_{x\to 0}(1+x)^\frac{1}{x}= e$. I get the answer to be 1.

$\displaystyle (\frac{1}{x})(\ln{1+x})= \infty\cdot 0 = 0 $

so $\displaystyle e^0 = 1$

What am I doing wrong? Thanks

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- Aug 20th 2008, 01:05 PM11rdc11help with limit
I'm trying to figure out how the $\displaystyle \lim_{x\to 0}(1+x)^\frac{1}{x}= e$. I get the answer to be 1.

$\displaystyle (\frac{1}{x})(\ln{1+x})= \infty\cdot 0 = 0 $

so $\displaystyle e^0 = 1$

What am I doing wrong? Thanks - Aug 20th 2008, 01:16 PMwingless
$\displaystyle \infty\cdot 0$ is an indeterminate form.

See Indeterminate form - Wikipedia, the free encyclopedia. - Aug 20th 2008, 01:35 PM11rdc11
Thanks and i just covered that section lol. I just made the assumption that 0 multiplied by anything is 0.

- Aug 20th 2008, 01:46 PMgalactus
One way is to build on the differentiability of ln(x).

More specifically,on the derivative of ln(x) at x=1.

$\displaystyle \frac{d}{dx}[ln(x)]\left|=\frac{1}{x}\right|_{x=1}=1$

If we express this using the definition of a derivative, we get:

$\displaystyle 1=\lim_{h\to 0}\frac{ln(1+h)-ln(1)}{h}=\lim_{h\to 0}\frac{ln(1+h)}{h}=\lim_{h\to 0}ln(1+h)^{\frac{1}{h}}$

And it follows that we get:

$\displaystyle \displaystyle{e=e^{\left(\displaystyle{\lim_{h\to 0}ln(1+h)^{\frac{1}{h}}}\right)}}$

Now, from the continuity of e, we can write:

$\displaystyle \displaystyle{e=\lim_{h\to 0}e^{ln(1+h)^{\frac{1}{h}}}=\lim_{h\to 0}(1+h)^{\frac{1}{h}}}$