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

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

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

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

0

Solution steps

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

Simplify

e^{- n} : \frac{1}{e ^{n}}

= 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

= \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

= \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

\frac{\partial}{\partial y} (2 x + 2 y)

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

y = \frac{5}{x}

y^{^{''}} + 6 y^{^{'}} + 9 = 0

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