What is the limit of $x^{\frac{1}{x}}$ $x^(1/x)$ as $x \to \infty$ $x->oo$ ?

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What is the limit of $x^{\frac{1}{x}}$ $x^(1/x)$ as $x \to \infty$ $x->oo$ ?

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Answer 1 · 2017-02-24T17:00:08

$L t_{x \to \infty} x^{\frac{1}{x}} = 1$ $Lt_(x->oo)x^(1/x)=1$

Explanation:

Let $x^{\frac{1}{x}} = e^{a}$ $x^(1/x)=e^a$ , then taking natural log on both sides, we get

$\frac{1}{x} ln x = a$ $1/xlnx=a$ and $x^{\frac{1}{x}} = e^{\frac{ln x}{x}}$ $x^(1/x)=e^(lnx/x)$

Hence, $L t_{x \to \infty} x^{\frac{1}{x}} = L t_{x \to \infty} e^{\frac{ln x}{x}}$ $Lt_(x->oo)x^(1/x)=Lt_(x->oo)e^(lnx/x)$

and as $e^{u}$ $e^u$ is continuous $L t_{x \to \infty} x^{\frac{1}{x}} = e_{x \to \infty}^{L t_{x \to \infty} (\frac{ln x}{x})}$ $Lt_(x->oo)x^(1/x)=e^(Lt_(x->oo)(lnx/x))$

Now in $L t_{x \to \infty} e^{\frac{ln x}{x}}$ $Lt_(x->oo)e^(lnx/x)$ as $x \to \infty$ $x->oo$ , both numerator and denominators tend to infinity,

therefore we can apply L'Hospital's Rule and

$L t_{x \to \infty} (\frac{ln x}{x}) = L t_{x \to \infty} (\frac{\frac{1}{x}}{1}) = L t_{x \to \infty} \frac{1}{x} = 0$ $Lt_(x->oo)(lnx/x)=Lt_(x->oo)((1/x)/1)=Lt_(x->oo)1/x=0$

Hence $L t_{x \to \infty} x^{\frac{1}{x}} = e^{0} = 1$ $Lt_(x->oo)x^(1/x)=e^0=1$

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