Limit problem involving x to the power of a function

  • Thread starter Wormaldson
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Homework Statement



Calculate the limit [tex]\lim_{x\rightarrow +\infty} x^{\frac{1}{x^2}}[/tex]

2. The attempt at a solution

At first I was tempted to rewrite the function as a composition and go from there. I soon realised that this wouldn't work as the term being raised to the power of a function of x was a different function of x; to my knowledge, decomposing functions in this way will only work for functions like [tex]a^{f(x)}[/tex]So no luck there.

The next thing I tried was manipulating the function algebraically; unfortunately I just ended up going in circles and wasn't able to get the function into a more useful form.

I came up with another method that seems superficially plausible, but I'm not sure if it's actually valid: that is, rewriting the problem as [tex]x^{\lim_{x\rightarrow +\infty} {\frac{1}{x^2}}}[/tex]Intuitively, this seems sensible; the limit at infinity of the exponent is 0 and any x > 0 raised to the power of 0 is 1. The course book mentions that you can take a limit "inside" a continuous function (the exact wording). Obviously the function in question isn't continuous on it's entire domain, but it is continuous for all x > 0. Unfortunately, however, the course book doesn't go into any greater detail, and the textbook doesn't address limits of this type at all. If taking the limit "inside" the function in such a manner is a valid operation, then I guess my problem is solved. If not, then I could use a suggestion as to what to try next.

As always, any help is much appreciated.
 

Answers and Replies

  • #2
LCKurtz
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If ##y = x^\frac 1 {x^2}##, try taking the limit of ##\ln y##.
 
  • #3
ehild
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Using that x=eln(x), the function can be written in the form [tex](e^{\ln(x)}) ^{\frac{1}{x^2}}=e^{\ln(x)\frac{1}{x^2}}[/tex]

ehild
 

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