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

The limit as t approaches infinity of (e^t+t^2)/(2e^t-t) is 1/2

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limit as t approaches infinity of (e^t+t^2)/(2e^t-t)

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

\frac{1}{2}

Decimal

0.5

Solution steps

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

Divide by

e^{t} : \frac{1 + \frac{t ^{2}}{e ^{t}}}{2 - \frac{t}{e ^{t}}}

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

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

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

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

= \frac{1}{2}

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