What is the limit as x approaches 0+of x/(e^{-1/x)} ?

The limit as x approaches 0+of x/(e^{-1/x)} is infinity

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limit as x approaches 0+of x/(e^{-1/x)}

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

\infty

Solution steps

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

Simplify

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

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

Apply the Limit Chain Rule:

\infty

= \infty

x \to 2 lim (6 x^{2} + 3 x - 9)

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

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

x \to - \infty lim (6 x^{3} - 5 x + 3)

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

What is the limit as x approaches 0+of x/(e^{-1/x)} ?
The limit as x approaches 0+of x/(e^{-1/x)} is infinity