What is the limit as x approaches 0+of x/(e^{-5/x)} ?

The limit as x approaches 0+of x/(e^{-5/x)} is infinity

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limit as x approaches 0+of x/(e^{-5/x)}

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

\infty

Solution steps

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

Simplify

\frac{x}{e ^{- \frac{5}{x}}} : e^{\frac{5}{x}} x

= x \to 0 + lim (e^{\frac{5}{x}} x)

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to 0 + lim \frac{- \frac{5 e ^{\frac{5}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to 0 + lim \frac{- \frac{5 e ^{\frac{5}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{5 e ^{\frac{5}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 5 e^{\frac{5}{x}}

= x \to 0 + lim (5 e^{\frac{5}{x}})

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

= 5 \cdot x \to 0 + lim (e^{\frac{5}{x}})

Apply the Limit Chain Rule:

\infty

= 5 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to 8 + lim (\frac{x + 5}{x - 8})

t \to 0 lim (\frac{5 ^{t} - 6 ^{t}}{t})

x \to 0 lim (x + 2 - 1)

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

x \to 0 lim ((2 - cos (x))^{c s c^{2} (x)})

What is the limit as x approaches 0+of x/(e^{-5/x)} ?
The limit as x approaches 0+of x/(e^{-5/x)} is infinity