What is the limit as x approaches 0-of-x^{-2}e^{1/x} ?

The limit as x approaches 0-of-x^{-2}e^{1/x} is 0

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limit as x approaches 0-of-x^{-2}e^{1/x}

x \to 0 - lim (- x^{- 2} e^{\frac{1}{x}})

0

Solution steps

x \to 0 - lim (- x^{- 2} e^{\frac{1}{x}})

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

= - x \to 0 - lim (x^{- 2} e^{\frac{1}{x}})

Rewrite for L'Hopital

= - x \to 0 - lim (\frac{x ^{- 2}}{\frac{1}{e ^{\frac{1}{x}}}})

Apply L'Hopital's Rule

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

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

Simplify

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

= - x \to 0 - lim (- \frac{2}{e ^{- \frac{1}{x}} x})

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

= - (- 2 \cdot x \to 0 - lim (\frac{1}{e ^{- \frac{1}{x}} x}))

x \to a lim [\frac{f ( x )}{g ( x )}] = \frac{lim _{x \to a} f ( x )}{lim _{x \to a} g ( x )}, x \to a lim g (x) \neq = 0

With the exception of indeterminate form

= - - 2 \cdot \frac{lim _{x \to 0 -} ( 1 )}{lim _{x \to 0 -} ( e ^{- \frac{1}{x}} x )}

x \to 0 - lim (1) = 1

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

= - (- 2 \cdot \frac{1}{- \infty})

Simplify

- (- 2 \cdot \frac{1}{- \infty}) : 0

= 0

t \to - 4 lim (\frac{t ^{2}}{4 - t})

x \to \frac{6}{7} lim (\frac{1}{7 x - 6})

x \to - 3 + lim (\frac{x ^{3} + 9}{x + 3})

x \to 0 lim (x^{0.5})

x \to 7 - lim (ln (7 - x))

What is the limit as x approaches 0-of-x^{-2}e^{1/x} ?
The limit as x approaches 0-of-x^{-2}e^{1/x} is 0