What is the limit as n approaches infinity of n^2e^{-3n} ?

The limit as n approaches infinity of n^2e^{-3n} is 0

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limit as n approaches infinity of n^2e^{-3n}

n \to \infty lim (n^{2} e^{- 3 n})

0

Solution steps

n \to \infty lim (n^{2} e^{- 3 n})

Rewrite for L'Hopital

= n \to \infty lim (\frac{n ^{2}}{\frac{1}{e ^{- 3 n}}})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{2 n}{e ^{3 n} \cdot 3})

= n \to \infty lim (\frac{2 n}{e ^{3 n} \cdot 3})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{2}{9 e ^{3 n}})

= n \to \infty lim (\frac{2}{9 e ^{3 n}})

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

= \frac{2}{9} \cdot n \to \infty lim (\frac{1}{e ^{3 n}})

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{2}{9} \cdot \frac{lim _{n \to \infty} ( 1 )}{lim _{n \to \infty} ( e ^{3 n} )}

n \to \infty lim (1) = 1

n \to \infty lim (e^{3 n}) = \infty

= \frac{2}{9} \cdot \frac{1}{\infty}

Simplify

\frac{2}{9} \cdot \frac{1}{\infty} : 0

= 0

\frac{( s - 3 )}{( s ^{2} + 4 s + 13 )}

y^{^{'}} - 7 y = - x

y^{^{'}} = \frac{3100}{1 + 0.25 t}

\int (5 x + 4)^{5} d x

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What is the limit as n approaches infinity of n^2e^{-3n} ?
The limit as n approaches infinity of n^2e^{-3n} is 0