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
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
$\displaystyle \infty\cdot 0$ is an indeterminate form.
See Indeterminate form - Wikipedia, the free encyclopedia.
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:
$\displaystyle \displaystyle{e=\lim_{h\to 0}e^{ln(1+h)^{\frac{1}{h}}}=\lim_{h\to 0}(1+h)^{\frac{1}{h}}}$