Limit problem involving x to the power of a function

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SUMMARY

The limit problem involving the expression \(\lim_{x\rightarrow +\infty} x^{\frac{1}{x^2}}\) can be effectively solved by taking the natural logarithm of the function. By rewriting the function as \(y = x^{\frac{1}{x^2}}\) and applying the limit to \(\ln y\), the expression transforms into \(e^{\ln(x)\frac{1}{x^2}}\). As \(x\) approaches infinity, \(\ln(x)\) grows slower than \(x^2\), leading to a limit of 0 for the exponent, thus confirming that the original limit evaluates to 1.

PREREQUISITES
  • Understanding of limits in calculus
  • Familiarity with natural logarithms and their properties
  • Knowledge of exponential functions
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the properties of limits involving exponential functions
  • Learn about the application of L'Hôpital's Rule for indeterminate forms
  • Explore the concept of continuity in functions and its implications for limits
  • Investigate the behavior of logarithmic functions as their arguments approach infinity
USEFUL FOR

Students studying calculus, particularly those focusing on limits and exponential functions, as well as educators seeking to clarify concepts related to limit evaluation techniques.

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Homework Statement



Calculate the limit \lim_{x\rightarrow +\infty} x^{\frac{1}{x^2}}

2. The attempt at a solution

At first I was tempted to rewrite the function as a composition and go from there. I soon realized 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 a^{f(x)}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 x^{\lim_{x\rightarrow +\infty} {\frac{1}{x^2}}}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.
 
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If ##y = x^\frac 1 {x^2}##, try taking the limit of ##\ln y##.
 
Using that x=eln(x), the function can be written in the form (e^{\ln(x)}) ^{\frac{1}{x^2}}=e^{\ln(x)\frac{1}{x^2}}

ehild
 

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