What is the limit as x approaches infinity of 9x^2e^{-x} ?

The limit as x approaches infinity of 9x^2e^{-x} is 0

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limit as x approaches infinity of 9x^2e^{-x}

x \to \infty lim (9 x^{2} e^{- x})

0

Solution steps

x \to \infty lim (9 x^{2} e^{- x})

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

= 9 \cdot x \to \infty lim (x^{2} e^{- x})

Rewrite for L'Hopital

= 9 \cdot x \to \infty lim (\frac{x ^{2}}{\frac{1}{e ^{- x}}})

Apply L'Hopital's Rule

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

= 9 \cdot x \to \infty lim (\frac{2 x}{e ^{x}})

Apply L'Hopital's Rule

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

= 9 \cdot x \to \infty lim (\frac{2}{e ^{x}})

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

= 9 \cdot 2 \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

= 9 \cdot 2 \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

= 9 \cdot 2 \cdot \frac{1}{\infty}

Simplify

9 \cdot 2 \cdot \frac{1}{\infty} : 0

= 0

(\frac{x ^{3}}{6} + \frac{1}{2 x})^{^{'}}

\int_{6}^{\infty} \frac{1}{( x - 5 ) ^{\frac{3}{2}}} d x

\int_{0}^{5} x^{2} + 25 d x

y = sin (x y)

\frac{\partial}{\partial t} (x ln (t))

What is the limit as x approaches infinity of 9x^2e^{-x} ?
The limit as x approaches infinity of 9x^2e^{-x} is 0