What is the limit as x approaches 0+of x/(e^{-9/x)} ?

The limit as x approaches 0+of x/(e^{-9/x)} is infinity

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limit as x approaches 0+of x/(e^{-9/x)}

x \to 0 + lim (\frac{x}{e ^{- \frac{9}{x}}})

\infty

Solution steps

x \to 0 + lim (\frac{x}{e ^{- \frac{9}{x}}})

Simplify

\frac{x}{e ^{- \frac{9}{x}}} : e^{\frac{9}{x}} x

= x \to 0 + lim (e^{\frac{9}{x}} x)

Rewrite for L'Hopital

= x \to 0 + lim (\frac{e ^{\frac{9}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim \frac{- \frac{9 e ^{\frac{9}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to 0 + lim \frac{- \frac{9 e ^{\frac{9}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{9 e ^{\frac{9}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 9 e^{\frac{9}{x}}

= x \to 0 + lim (9 e^{\frac{9}{x}})

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

= 9 \cdot x \to 0 + lim (e^{\frac{9}{x}})

Apply the Limit Chain Rule:

\infty

= 9 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to 0 + lim (tan (x^{x}))

x \to 1 lim ((x^{2} - 4) (x - 3))

x \to 0 - lim ((x - 5) e^{\frac{1}{x}})

x \to 0 lim (\frac{2 x}{x + 1})

x \to 1 lim (x^{2} - 3 x + 5)

What is the limit as x approaches 0+of x/(e^{-9/x)} ?
The limit as x approaches 0+of x/(e^{-9/x)} is infinity