What is the limit as x approaches-infinity of 6/((1+e^{-x))} ?

The limit as x approaches-infinity of 6/((1+e^{-x))} is 0

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limit as x approaches-infinity of 6/((1+e^{-x))}

x \to - \infty lim (\frac{6}{( 1 + e ^{- x} )})

0

Solution steps

x \to - \infty lim (\frac{6}{1 + e ^{- x}})

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

= 6 \cdot x \to - \infty lim (\frac{1}{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

= 6 \cdot \frac{lim _{x \to - \infty} ( 1 )}{lim _{x \to - \infty} ( 1 + e ^{- x} )}

x \to - \infty lim (1) = 1

x \to - \infty lim (1 + e^{- x}) = \infty

= 6 \cdot \frac{1}{\infty}

Simplify

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

= 0

x \to 7 lim (\frac{x ^{3} - 4 x ^{2} - 19 x - 14}{x ^{2} - 8 x - 7})

y^{^{''}} + 2 y^{^{'}} + y = x^{2} - 2 x + 1

\frac{d}{d x} ((x^{4} ln (x))^{5})

x \to 6 lim (ln (6 - x))

x \to 0 lim (x - \frac{1}{x})

What is the limit as x approaches-infinity of 6/((1+e^{-x))} ?
The limit as x approaches-infinity of 6/((1+e^{-x))} is 0