# limits

• Dec 13th 2005, 11:51 AM
bobby77
please see the attatchment
• Dec 13th 2005, 01:51 PM
mathproblem
As x gets indefinetly large 1/x will get closer and closer to zero and any non zero real number rasied to the zero power yeilds one so the lim as x-->infinity of (x)^(1/x) = 1
• Dec 13th 2005, 01:51 PM
CaptainBlack
Quote:

Originally Posted by bobby77
please see the attatchment

You are asking - what is:

$Lt_{x\rightarrow \infty} \sqrt[x] x$

with a choice of answers $\infty,\ 1,\ 0.$

Consider:

$log(Lt_{x\rightarrow \infty} \sqrt[x] x)$,

which is equal to:

$Lt_{x\rightarrow \infty}log( \sqrt[x] x)\ =\ Lt_{x\rightarrow \infty} \frac{log(x)}{x}$.

Now if you know that $log(x)$ grows more slowly than any
positive power of $x$ you are done, as:

$Lt_{x\rightarrow \infty} \frac{log(x)}{x}\ =\ 0$,

and so:

$Lt_{x\rightarrow \infty} \sqrt[x] x\ =\ 1$.

If you don't know that $log(x)$ grows more slowly than any
positive power of $x$, then you can get the same result by a slightly
convoluted application of L'Hopital rule.

RonL
• Dec 13th 2005, 01:56 PM
CaptainBlack
Quote:

Originally Posted by mathproblem
As x gets indefinetly large 1/x will get closer and closer to zero and any non zero real number rasied to the zero power yeilds one so the lim as x-->infinity of (x)^(1/x) = 1

Would not the same argument apply to $\sqrt[x]{x^x}$?

RonL