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}})

Simplify

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

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

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

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

Apply the common limit:

x \to - \infty lim (e^{x}) = 0

= - (- 0)

Simplify

= 0

y = x^{2} - 4 x, y = 2 x - x^{2}

f (x) = x + \frac{1}{x}, at x = 3

\frac{4 x ^{2} + 1}{x ^{2} + 5}

\frac{d}{d x} (x^{4} - 6 x^{2} + 4)

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

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