# Thread: need a little limit help action

1. ## need a little limit help action

It can be shown that .
Use this limit to evaluate the limits below.

(a) =

(b) =

need some direction/help...thanks

2. For the first one, define $u=\frac1x,$ so

$\lim_{x\to0}(1+x)^{1/x}=\lim_{u\to\infty}\left(1+\frac1u\right)^u=e.$

For the second one, you can take it more generally:

$\forall x\in\mathbb R,\,\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x.$

3. thanks for the help, i know these are right, however when i try to submit them, it tells me they are wrong...do you have any idea why...

thanks

4. Originally Posted by mer1988
It can be shown that .
Use this limit to evaluate the limits below.

(a) =

(b) =

need some direction/help...thanks
the previous solution was also the first thing that came into my mind, yet your teacher didn't want it that way.. i haev another approach..
we know that $f(x)=e^{\ln {f(x)}}\,$ thus if we let $f(x)= (1+x)^{\frac{1}{x}}\,$, then we have
$e^{\ln {(1+x)^{\frac{1}{x}}}} = e^{\frac{1}{x} \ln {(1+x)}}$, and take the limit as $x \rightarrow 0^+$ so it becomes
$\lim_{x \rightarrow 0^+}e^{\frac{1}{x} \ln {(1+x)}} = e^{\lim_{x \rightarrow 0^+}{\frac{1}{x} \ln {(1+x)}}} = e^{\lim_{x \rightarrow 0^+}{\frac{\ln {(1+x)}}{x}}}$
and the limit part is in the case of $\frac{0}{0}$, you may use L'hospital's Rule, so the limit part equals 1 and so $e^1 = e$..

you can do the same thing for 2..