What is the limit as x approaches 0+of x/(e^{-4/x)} ?

The limit as x approaches 0+of x/(e^{-4/x)} is infinity

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limit as x approaches 0+of x/(e^{-4/x)}

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

\infty

Solution steps

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

Simplify

\frac{x}{e ^{- \frac{4}{x}}} : e^{\frac{4}{x}} x

= x \to 0 + lim (e^{\frac{4}{x}} x)

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to 0 + lim \frac{- \frac{4 e ^{\frac{4}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to 0 + lim \frac{- \frac{4 e ^{\frac{4}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{4 e ^{\frac{4}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 4 e^{\frac{4}{x}}

= x \to 0 + lim (4 e^{\frac{4}{x}})

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

= 4 \cdot x \to 0 + lim (e^{\frac{4}{x}})

Apply the Limit Chain Rule:

\infty

= 4 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to 2 lim (cos (π x))

x \to 1 lim (\frac{cos ( π x )}{x})

x \to 0 lim (x^{3} - 12 x - 5)

x \to 0 lim (ln (cot (x)))

x \to 7 lim (4 x^{2} - 11 x + 3)

What is the limit as x approaches 0+of x/(e^{-4/x)} ?
The limit as x approaches 0+of x/(e^{-4/x)} is infinity