What is the limit as x approaches infinity of-xe^{1-x} ?

The limit as x approaches infinity of-xe^{1-x} 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-xe^{1-x}

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

0

Solution steps

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

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

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

Simplify

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

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

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

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

= - e \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

= - e \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

= - e \frac{1}{\infty}

Simplify

- e \frac{1}{\infty} : 0

= 0

\int \frac{7 r ^{2}}{r + 3} d r

2 x^{2} - 9 x + 10, at x = 1

\int \frac{( 1 - 2 y )}{y} d y

\int e^{2 x} \cdot sin (2 x) d x

f (x) = x^{2}, at x = - 5

What is the limit as x approaches infinity of-xe^{1-x} ?
The limit as x approaches infinity of-xe^{1-x} is 0