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

The limit as x approaches 0-of xe^{-1/x} is -infinity

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limit as x approaches 0-of xe^{-1/x}

x \to 0 - lim (x e^{- \frac{1}{x}})

- \infty

Solution steps

x \to 0 - lim (x e^{- \frac{1}{x}})

Rewrite for L'Hopital

= x \to 0 - lim (\frac{e ^{- \frac{1}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 - lim \frac{\frac{e ^{- \frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to 0 - lim \frac{\frac{e ^{- \frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{\frac{e ^{- \frac{1}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : - e^{- \frac{1}{x}}

= x \to 0 - lim (- e^{- \frac{1}{x}})

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

= - x \to 0 - lim (e^{- \frac{1}{x}})

Apply the Limit Chain Rule:

\infty

= - \infty

x \to 0 lim (\frac{6 ^{x} - 6 ^{- x}}{x})

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

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

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

x \to 0 lim (\frac{1}{x} + \frac{1}{e ^{x} - 1})

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