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

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

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limit as x approaches infinity of-2xe^{-2x}

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

0

Solution steps

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

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{1}{e ^{2 x} \cdot 2})

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

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

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

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

x \to \infty lim (1) = 1

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

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

Simplify

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

= 0

y = \frac{1}{2 cos ^{2} ( x )}

i = 0 \sum \infty \frac{i}{2 ^{i}}

\frac{d y}{d x} = \frac{x ^{3}}{y}

\int \frac{x}{( x + 1 ) ^{8}} d x

\int \frac{e ^{5 x}}{e ^{5 x} + 4} d x

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