please see the attatchment
You are asking - what is:Originally Posted by bobby77
$\displaystyle Lt_{x\rightarrow \infty} \sqrt[x] x$
with a choice of answers $\displaystyle \infty,\ 1,\ 0.$
Consider:
$\displaystyle log(Lt_{x\rightarrow \infty} \sqrt[x] x)$,
which is equal to:
$\displaystyle Lt_{x\rightarrow \infty}log( \sqrt[x] x)\ =\ Lt_{x\rightarrow \infty} \frac{log(x)}{x}$.
Now if you know that $\displaystyle log(x)$ grows more slowly than any
positive power of $\displaystyle x$ you are done, as:
$\displaystyle Lt_{x\rightarrow \infty} \frac{log(x)}{x}\ =\ 0$,
and so:
$\displaystyle Lt_{x\rightarrow \infty} \sqrt[x] x\ =\ 1$.
If you don't know that $\displaystyle log(x)$ grows more slowly than any
positive power of $\displaystyle x$, then you can get the same result by a slightly
convoluted application of L'Hopital rule.
RonL