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

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

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

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

\infty

Solution steps

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

Rewrite for L'Hopital

= x \to 0 + lim (\frac{- 1 + 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 0 lim ((1 + 4 x)^{- \frac{3}{x}})

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

x \to 0 lim (1^{\frac{l n ( 6 )}{x}})

x \to - 4 lim (∣ x + 4 ∣ - 2 x)

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

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