What is the limit as x approaches infinity of ((x^{13}))/(e^x) ?

The limit as x approaches infinity of ((x^{13}))/(e^x) is 0

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limit as x approaches infinity of ((x^{13}))/(e^x)

x \to \infty lim (\frac{( x ^{13} )}{e ^{x}})

0

Solution steps

x \to \infty lim (\frac{x ^{13}}{e ^{x}})

\frac{x ^{13}}{e ^{x}} = x^{13} \frac{1}{e ^{x}}

= x \to \infty lim (x^{13} \frac{1}{e ^{x}})

Simplify

x^{13} \frac{1}{e ^{x}} : \frac{x ^{13}}{e ^{x}}

= x \to \infty lim (\frac{x ^{13}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{13 x ^{12}}{e ^{x}})

= x \to \infty lim (\frac{13 x ^{12}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{156 x ^{11}}{e ^{x}})

= x \to \infty lim (\frac{156 x ^{11}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{1716 x ^{10}}{e ^{x}})

= x \to \infty lim (\frac{1716 x ^{10}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{17160 x ^{9}}{e ^{x}})

= x \to \infty lim (\frac{17160 x ^{9}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{154440 x ^{8}}{e ^{x}})

= x \to \infty lim (\frac{154440 x ^{8}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{1235520 x ^{7}}{e ^{x}})

= x \to \infty lim (\frac{1235520 x ^{7}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{8648640 x ^{6}}{e ^{x}})

= x \to \infty lim (\frac{8648640 x ^{6}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{51891840 x ^{5}}{e ^{x}})

= x \to \infty lim (\frac{51891840 x ^{5}}{e ^{x}})

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

= 51891840 \cdot x \to \infty lim (\frac{x ^{5}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{5 x ^{4}}{e ^{x}})

= 51891840 \cdot x \to \infty lim (\frac{5 x ^{4}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{20 x ^{3}}{e ^{x}})

= 51891840 \cdot x \to \infty lim (\frac{20 x ^{3}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{60 x ^{2}}{e ^{x}})

= 51891840 \cdot x \to \infty lim (\frac{60 x ^{2}}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{120 x}{e ^{x}})

= 51891840 \cdot x \to \infty lim (\frac{120 x}{e ^{x}})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{120}{e ^{x}})

= 51891840 \cdot x \to \infty lim (\frac{120}{e ^{x}})

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

= 51891840 \cdot 120 \cdot x \to \infty lim (\frac{1}{e ^{x}})

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

= 51891840 \cdot 120 \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( e ^{x} )}

x \to \infty lim (1) = 1

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

= 51891840 \cdot 120 \cdot \frac{1}{\infty}

Simplify

51891840 \cdot 120 \cdot \frac{1}{\infty} : 0

= 0

x \to - 2 lim (3 x - \frac{6}{x})

y \to 0 lim (\frac{cos ( y )}{y})

x \to 0 lim ((9^{x} + x)^{\frac{2}{x}})

x \to 3 lim (3 x^{2} + x)

x \to 1 lim ([\frac{x ^{3} - 1}{x ^{2} - 1}])

What is the limit as x approaches infinity of ((x^{13}))/(e^x) ?
The limit as x approaches infinity of ((x^{13}))/(e^x) is 0