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

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

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

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

0

Solution steps

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

Simplify

e^{- n + 2} : e^{- n} e^{2}

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

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

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

Simplify

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

= e^{2} \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

= e^{2} \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

= e^{2} \frac{1}{\infty}

Simplify

e^{2} \frac{1}{\infty} : 0

= 0

h \to 0 lim (csc (h) - cot (h))

x \to - 3 lim ((4 x - 2 \cdot 3)^{2})

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

x \to 0 lim (3 x^{4} - 2 x - 1)

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