It can be shown that .
Use this limit to evaluate the limits below.
(a) =
(b) =
need some direction/help...thanks
For the first one, define $\displaystyle u=\frac1x,$ so
$\displaystyle \lim_{x\to0}(1+x)^{1/x}=\lim_{u\to\infty}\left(1+\frac1u\right)^u=e.$
For the second one, you can take it more generally:
$\displaystyle \forall x\in\mathbb R,\,\lim_{n\to\infty}\left(1+\frac xn\right)^n=e^x.$
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 $\displaystyle f(x)=e^{\ln {f(x)}}\,$ thus if we let $\displaystyle f(x)= (1+x)^{\frac{1}{x}}\,$, then we have
$\displaystyle e^{\ln {(1+x)^{\frac{1}{x}}}} = e^{\frac{1}{x} \ln {(1+x)}}$, and take the limit as $\displaystyle x \rightarrow 0^+$ so it becomes
$\displaystyle \lim_{x \rightarrow 0^+}e^{\frac{1}{x} \ln {(1+x)}} = e^{\lim_{x \rightarrow 0^+}{\frac{1}{x} \ln {(1+x)}}} = e^{\lim_{x \rightarrow 0^+}{\frac{\ln {(1+x)}}{x}}}$
and the limit part is in the case of $\displaystyle \frac{0}{0}$, you may use L'hospital's Rule, so the limit part equals 1 and so$\displaystyle e^1 = e$..
you can do the same thing for 2..