need a little limit help action

**mer1988** · October 30th 2007, 05:31 PM

It can be shown that

.
Use this limit to evaluate the limits below.

(a)

=

(b)

=

need some direction/help...thanks

**Krizalid** · October 30th 2007, 06:16 PM

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:

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

**mer1988** · October 30th 2007, 09:05 PM

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

**kalagota** · October 30th 2007, 09:34 PM

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..

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