What is the limit as x approaches infinity of (x^2-1)e^{-x^2} ?

The limit as x approaches infinity of (x^2-1)e^{-x^2} is 0

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limit as x approaches infinity of (x^2-1)e^{-x^2}

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

0

Solution steps

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

Rewrite for L'Hopital

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{2 x}{e ^{x^{2}} \cdot 2 x})

= x \to \infty lim (\frac{2 x}{e ^{x^{2}} \cdot 2 x})

Apply L'Hopital's Rule

: x \to \infty lim (\frac{2}{2 ( 2 e ^{x^{2}} x ^{2} + e ^{x^{2}} )})

= x \to \infty lim (\frac{2}{2 ( 2 e ^{x^{2}} x ^{2} + e ^{x^{2}} )})

Simplify

= x \to \infty lim (\frac{1}{2 e ^{x^{2}} x ^{2} + e ^{x^{2}}})

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

= \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( 2 e ^{x^{2}} x ^{2} + e ^{x^{2}} )}

x \to \infty lim (1) = 1

x \to \infty lim (2 e^{x^{2}} x^{2} + e^{x^{2}}) = \infty

= \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

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