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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

g (x) = x (x - 4)^{3}

x \to 0 lim (\frac{7 x ^{4} - x ^{3}}{x})

\int e^{t x} d x

\int 4 e^{3 x} d x

4 y^{^{''}} - y^{^{'}} + 1 = 0

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