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

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

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

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

\infty

Solution steps

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 - 7 - lim (f (x))

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

x \to 4 lim (x + 5)

x \to 0 lim ((1 + 5 x)^{\frac{7}{x}})

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

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