# limits

• Dec 13th 2005, 10:51 AM
bobby77
• Dec 13th 2005, 12:51 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, 12:51 PM
CaptainBlack
Quote:

Originally Posted by bobby77

You are asking - what is:

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

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
• Dec 13th 2005, 12:56 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 $\displaystyle \sqrt[x]{x^x}$?

RonL