What is the limit as x approaches 0+of x/(e^{-3/x)} ?

The limit as x approaches 0+of x/(e^{-3/x)} is infinity

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limit as x approaches 0+of x/(e^{-3/x)}

x \to 0 + lim (\frac{x}{e ^{- \frac{3}{x}}})

\infty

Solution steps

x \to 0 + lim (\frac{x}{e ^{- \frac{3}{x}}})

Simplify

\frac{x}{e ^{- \frac{3}{x}}} : e^{\frac{3}{x}} x

= x \to 0 + lim (e^{\frac{3}{x}} x)

Rewrite for L'Hopital

= x \to 0 + lim (\frac{e ^{\frac{3}{x}}}{\frac{1}{x}})

Apply L'Hopital's Rule

: x \to 0 + lim \frac{- \frac{3 e ^{\frac{3}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

= x \to 0 + lim \frac{- \frac{3 e ^{\frac{3}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}}

Simplify

\frac{- \frac{3 e ^{\frac{3}{x}}}{x ^{2}}}{- \frac{1}{x ^{2}}} : 3 e^{\frac{3}{x}}

= x \to 0 + lim (3 e^{\frac{3}{x}})

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

= 3 \cdot x \to 0 + lim (e^{\frac{3}{x}})

Apply the Limit Chain Rule:

\infty

= 3 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

x \to 1 lim (\frac{[ x ] + 2}{3})

x \to π - lim (ln (x) sin (x))

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

x \to 1 lim (∣ (3 - 4 x) ∣)

x \to - \frac{π}{2} lim (- 4 x - \frac{π}{2})

What is the limit as x approaches 0+of x/(e^{-3/x)} ?
The limit as x approaches 0+of x/(e^{-3/x)} is infinity