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

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

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limit as x approaches infinity of xe^{-x}ln(x)

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

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

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

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

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

x \to \infty lim (1) = 1

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

= \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

x \to 0 lim (3 x^{4} - 2 x - 1)

x \to 0 + lim (x tan (\frac{π}{2} - x))

x \to 0 lim ((1 - sin (3 x))^{\frac{1}{2 x}})

x \to 4 + lim (x^{2} + 5 x - 35)

x \to \infty lim (3 (x))

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