What is the limit as t approaches infinity of (0.2t)/(t^2+1) ?

The limit as t approaches infinity of (0.2t)/(t^2+1) is 0

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limit as t approaches infinity of (0.2t)/(t^2+1)

t \to \infty lim (\frac{0.2 t}{t ^{2} + 1})

0

Solution steps

t \to \infty lim (\frac{0.2 t}{t ^{2} + 1})

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

= 0.2 \cdot t \to \infty lim (\frac{t}{t ^{2} + 1})

Divide by highest denominator power:

\frac{\frac{1}{t}}{1 + \frac{1}{t ^{2}}}

= 0.2 \cdot t \to \infty lim (\frac{\frac{1}{t}}{1 + \frac{1}{t ^{2}}})

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

= 0.2 \cdot \frac{lim _{t \to \infty} ( \frac{1}{t} )}{lim _{t \to \infty} ( 1 + \frac{1}{t ^{2}} )}

t \to \infty lim (\frac{1}{t}) = 0

t \to \infty lim (1 + \frac{1}{t ^{2}}) = 1

= 0.2 \cdot \frac{0}{1}

Simplify

0.2 \cdot \frac{0}{1} : 0

= 0

\frac{d}{d x} (2 (- 2 e^{- x^{2}} x^{2} + e^{- x^{2}}))

x \to 5 lim (2 x^{3} - 3 x^{2} - x - 4)

t y^{^{'}} + y = t

f (x) = \frac{6 x ^{5}}{5} - x + 3 e^{x}

x, at 36

What is the limit as t approaches infinity of (0.2t)/(t^2+1) ?
The limit as t approaches infinity of (0.2t)/(t^2+1) is 0