What is the limit as x approaches infinity of ((x^p)/(e^x))^{1/n} ?

The limit as x approaches infinity of ((x^p)/(e^x))^{1/n} is 0

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limit as x approaches infinity of ((x^p)/(e^x))^{1/n}

x \to \infty lim ((\frac{x ^{p}}{e ^{x}})^{\frac{1}{n}})

0

Solution steps

x \to \infty lim ((\frac{x ^{p}}{e ^{x}})^{\frac{1}{n}})

x \to a lim [f (x)]^{b} = [x \to a lim f (x)]^{b}

With the exception of indeterminate form

= (x \to \infty lim (\frac{x ^{p}}{e ^{x}}))^{\frac{1}{n}}

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{p}}{e ^{x}} = (x^{p}) e^{- x}

= (x \to \infty lim (x^{p} e^{- x}))^{\frac{1}{n}}

Rewrite for L'Hopital

= (x \to \infty lim (\frac{x ^{p}}{\frac{1}{e ^{- x}}}))^{\frac{1}{n}}

Apply L'Hopital's Rule

= (x \to \infty lim (\frac{p x ^{p - 1}}{e ^{x}}))^{\frac{1}{n}}

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= (p \cdot x \to \infty lim (\frac{x ^{p - 1}}{e ^{x}}))^{\frac{1}{n}}

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= (p \frac{lim _{x \to \infty} ( x ^{p - 1} )}{lim _{x \to \infty} ( e ^{x} )})^{\frac{1}{n}}

x \to \infty lim (x^{p - 1}) =

Undefined

x \to \infty lim (e^{x}) = \infty

= (p \frac{Undefined}{\infty})^{\frac{1}{n}}

Simplify

(p \frac{Undefined}{\infty})^{\frac{1}{n}} : 0

= 0

What is the limit as x approaches infinity of ((x^p)/(e^x))^{1/n} ?
The limit as x approaches infinity of ((x^p)/(e^x))^{1/n} is 0