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

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

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limit as x approaches infinity+of e^{-x}

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

0

Solution steps

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

Simplify

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

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

x \to \infty lim (1) = 1

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

= \frac{1}{\infty}

Apply Infinity Property:

\frac{c}{\infty} = 0

= 0

\int sin (y) - y sin (x) d x

\frac{\partial}{\partial x} (e^{x y^{2}} + 4 z y^{3})

(1 + x^{2}) \frac{d y}{d x} + x y = 0

y = x^{- \frac{1}{2}}

\frac{d}{d x} (x^{2} \cdot y^{2})

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