What is the limit as x approaches infinity of (x^2)/(e^{4x)} ?

The limit as x approaches infinity of (x^2)/(e^{4x)} is 0

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limit as x approaches infinity of (x^2)/(e^{4x)}

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

0

Solution steps

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

\frac{x ^{2}}{e ^{4 x}} = (x^{2}) e^{- 4 x}

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

Rewrite for L'Hopital

= x \to \infty lim (\frac{x ^{2}}{\frac{1}{e ^{- 4 x}}})

Apply L'Hopital's Rule

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

Apply L'Hopital's Rule

= x \to \infty lim (\frac{2}{16 e ^{4 x}})

Cancel the common factor:

2

= x \to \infty lim (\frac{1}{8 e ^{4 x}})

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

= \frac{1}{8} \cdot x \to \infty lim (\frac{1}{e ^{4 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

= \frac{1}{8} \cdot \frac{lim _{x \to \infty} ( 1 )}{lim _{x \to \infty} ( e ^{4 x} )}

x \to \infty lim (1) = 1

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

= \frac{1}{8} \cdot \frac{1}{\infty}

Simplify

\frac{1}{8} \cdot \frac{1}{\infty} : 0

= 0

What is the limit as x approaches infinity of (x^2)/(e^{4x)} ?
The limit as x approaches infinity of (x^2)/(e^{4x)} is 0