What is the limit as x approaches 0+of x/(e^{-8/x)} ?

The limit as x approaches 0+of x/(e^{-8/x)} is infinity

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limit as x approaches 0+of x/(e^{-8/x)}

x \to 0 + lim (\frac{x}{e ^{- \frac{8}{x}}})

\infty

Solution steps

x \to 0 + lim (\frac{x}{e ^{- \frac{8}{x}}})

Simplify

\frac{x}{e ^{- \frac{8}{x}}} : e^{\frac{8}{x}} x

= x \to 0 + lim (e^{\frac{8}{x}} x)

Rewrite for L'Hopital

= x \to 0 + lim (\frac{e ^{\frac{8}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim \frac{- \frac{8 e ^{\frac{8}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to 0 + lim \frac{- \frac{8 e ^{\frac{8}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{8 e ^{\frac{8}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 8 e^{\frac{8}{x}}

= x \to 0 + lim (8 e^{\frac{8}{x}})

x \to a lim [c \cdot f (x)] = c \cdot x \to a lim f (x)

= 8 \cdot x \to 0 + lim (e^{\frac{8}{x}})

Apply the Limit Chain Rule:

\infty

= 8 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to - 1 lim (\frac{x ^{2} + 6}{x + 1})

x \to - \infty lim (x, - 1)

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

x \to - \infty lim (\frac{6 x}{x + cos ( x )})

x \to 3 lim (x^{2} - 6 x + 4)

What is the limit as x approaches 0+of x/(e^{-8/x)} ?
The limit as x approaches 0+of x/(e^{-8/x)} is infinity