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

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

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

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

1

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

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

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

Apply the Limit Chain Rule:

1

= 1

\int \frac{1 - e ^{x}}{e ^{x} + 1} d x

g (x) = (\frac{9 + 6 x}{7 - 4 x})

\frac{\partial}{\partial z} (\frac{y}{z})

\frac{x}{1 + x}

\frac{d}{d x} ((A x^{2} + B x) e^{- x})

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