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}})

Apply the Limit Chain Rule:

\infty

= \infty

x^{3} + x^{2} + 5 x - 1

\int 3^{2 x} e^{x} d x

x \to - 4 lim (∣ x ∣)

\frac{d}{d x} (ln (x^{3} + 4))

h (x) = ln (\frac{x ^{2} - 1}{x ^{2} + 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