What is the limit as t approaches infinity of (98)/(1+4e^{-0.1(t))} ?

The limit as t approaches infinity of (98)/(1+4e^{-0.1(t))} is 98

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limit as t approaches infinity of (98)/(1+4e^{-0.1(t))}

t \to \infty lim (\frac{98}{1 + 4 e ^{- 0.1 (t)}})

98

Solution steps

t \to \infty lim (\frac{98}{1 + 4 e ^{- 0.1 t}})

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

= 98 \cdot t \to \infty lim (\frac{1}{1 + 4 e ^{- 0.1 t}})

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

= 98 \cdot \frac{lim _{t \to \infty} ( 1 )}{lim _{t \to \infty} ( 1 + 4 e ^{- 0.1 t} )}

t \to \infty lim (1) = 1

t \to \infty lim (1 + 4 e^{- 0.1 t}) = 1

= 98 \cdot \frac{1}{1}

Simplify

= 98

\int \frac{1}{10 x} d x

y^{2}

f^{^{'}} (x) = 3 arctan (x^{2} - 3 x + 2)

\int 1 - 4 w^{2} d w

\int \frac{sin ( 2 x )}{cos ^{3} ( 2 x )} d x

What is the limit as t approaches infinity of (98)/(1+4e^{-0.1(t))} ?
The limit as t approaches infinity of (98)/(1+4e^{-0.1(t))} is 98