What is the limit as x approaches infinity of-xe^{-x} ?

The limit as x approaches infinity of-xe^{-x} is 0

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limit as x approaches infinity of-xe^{-x}

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

0

Solution steps

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

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

= - 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

= - \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

= - \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

y^{^{''}} + 2 y^{^{'}} + y = x^{2}, y (0) = 1, y^{^{'}} (0) = 2

y^{^{''}} - y = 0, y (0) = 3, y^{^{'}} (0) = - 1

\frac{\partial}{\partial θ} (r (θ) θ)

x^{2}, - 1, 1, - 2 x^{4}

x ln (x)

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