What is the limit as n approaches infinity of 5ne^{-n} ?

The limit as n approaches infinity of 5ne^{-n} is 0

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limit as n approaches infinity of 5ne^{-n}

n \to \infty lim (5 n e^{- n})

0

Solution steps

n \to \infty lim (5 n e^{- n})

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

= 5 \cdot n \to \infty lim (n e^{- n})

Rewrite for L'Hopital

= 5 \cdot n \to \infty lim (\frac{n}{\frac{1}{e ^{- n}}})

Apply L'Hopital's Rule

: n \to \infty lim (\frac{1}{e ^{n}})

= 5 \cdot n \to \infty lim (\frac{1}{e ^{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

= 5 \cdot \frac{lim _{n \to \infty} ( 1 )}{lim _{n \to \infty} ( e ^{n} )}

n \to \infty lim (1) = 1

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

= 5 \cdot \frac{1}{\infty}

Simplify

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

= 0

y = (1 - e^{x})^{2}

y^{^{'}} = ry + k

\frac{d}{d x} (\frac{1}{( 2 - x ) ^{2}})

\int \frac{2 x - 4 + 6 x}{x} d x

4 x^{2} - 7 x + 5

What is the limit as n approaches infinity of 5ne^{-n} ?
The limit as n approaches infinity of 5ne^{-n} is 0