Question:
$\lim_{x\to\infty} x^{\frac{1}{x}} = $
$\lim_{x\to\infty} x^{\frac{1}{x}} = $
Updated On: May 12, 2024
- 1
- $\infty$
- 0
- $none\, of \, these$
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The Correct Option is A
Solution and Explanation
Let $ y = \lim_{x\to\infty} x^{\frac{1}{x}}$ .....(i)
Taking log in (i) on both sides, we get
$ \lim _{x\to \infty } \frac{1}{x } \log x $
Applying L'-Hospital's Rule, we get
$ \log y = \lim _{x\to \infty } \frac{\frac{1}{x}}{1} = \lim _{x\to \infty } \frac{1}{x} =0$
or $y = e^0 = 1$
Taking log in (i) on both sides, we get
$ \lim _{x\to \infty } \frac{1}{x } \log x $
Applying L'-Hospital's Rule, we get
$ \log y = \lim _{x\to \infty } \frac{\frac{1}{x}}{1} = \lim _{x\to \infty } \frac{1}{x} =0$
or $y = e^0 = 1$
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