# help with limit

• August 20th 2008, 01:05 PM
11rdc11
help with limit
I'm trying to figure out how the $\lim_{x\to 0}(1+x)^\frac{1}{x}= e$. I get the answer to be 1.

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

so $e^0 = 1$

What am I doing wrong? Thanks
• August 20th 2008, 01:16 PM
wingless
$\infty\cdot 0$ is an indeterminate form.

See Indeterminate form - Wikipedia, the free encyclopedia.
• August 20th 2008, 01:35 PM
11rdc11
Thanks and i just covered that section lol. I just made the assumption that 0 multiplied by anything is 0.
• August 20th 2008, 01:46 PM
galactus
One way is to build on the differentiability of ln(x).

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

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

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

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

And it follows that we get:

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

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

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