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

The limit as x approaches infinity of x^{-2}e^{2x} is infinity

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

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

\infty

Solution steps

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

Simplify

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

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

Apply L'Hopital's Rule

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

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

Apply L'Hopital's Rule

: x \to \infty lim (\frac{4 e ^{2 x}}{2})

= x \to \infty lim (\frac{4 e ^{2 x}}{2})

Simplify

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

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

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

Apply the Limit Chain Rule:

\infty

= 2 \cdot \infty

Apply Infinity Property:

c \cdot \infty = \infty

= \infty

\frac{d}{d x} (e^{(2 x)^{3} - 2})

y^{^{'}} = \frac{( 4 t ^{3} + 1 )}{2 y - 6}

\int tan^{3} (x) sec^{2} (x) d x

y = 15 - x^{2}, y = x^{2} - 3

\int_{1}^{5} \frac{1}{1 + 2 x} d x

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