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Prove It
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#44
Dec22-10, 02:34 AM
P: 5
The standard method of solving indeterminate forms is to apply a transformation to get [tex]\displaystyle \frac{0}{0}[/tex] or [tex]\displaystyle \frac{\infty}{\infty}[/tex].

In this case [tex]\displaystyle \lim_{x \to \infty}x^{\frac{1}{x}} = \lim_{x \to \infty}e^{\ln{\left(x^{\frac{1}{x}}\right)}}[/tex]

[tex]\displaystyle = \lim_{x \to \infty}e^{\frac{1}{x}\ln{x}}[/tex]

[tex]\displaystyle = \lim_{x \to \infty}e^{\frac{\ln{x}}{x}}[/tex]

[tex]\displaystyle = e^{\lim_{x \to \infty}\frac{\ln{x}}{x}}[/tex] from the continutiy of the exponential function, this also gives the required form [tex]\displaystyle \frac{\infty}{\infty}[/tex] so that you can use L'Hospital's Rule

[tex]\displaystyle = e^{\lim_{x \to \infty}\frac{\frac{1}{x}}{1}}[/tex] by L'Hospital's Rule

[tex]\displaystyle = e^{\lim_{x \to \infty}\frac{1}{x}}[/tex]

[tex]\displaystyle = e^0[/tex]

[tex]\displaystyle = 1[/tex].