Question

use lhôpitals rule to find the limit.
lim_{t→∞} (e^{5t}+t^{2})/(2e^{5t}-1)
lim_{t→∞} (e^{5t}+t^{2})/(2e^{5t}-1)=□ (type an exact answer.)

Explanation:

Step1: Check indeterminate form

As \(t
ightarrow\infty\), \(\lim_{t
ightarrow\infty}\frac{e^{5t}+t^{2}}{2e^{5t}-t}\) is in the \(\frac{\infty}{\infty}\) form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of \(e^{5t}+t^{2}\) is \(5e^{5t} + 2t\), and the derivative of \(2e^{5t}-t\) is \(10e^{5t}-1\). So we get \(\lim_{t
ightarrow\infty}\frac{5e^{5t}+2t}{10e^{5t}-1}\), which is still in \(\frac{\infty}{\infty}\) form.

Step3: Apply L'Hopital's Rule again

Differentiate the new - formed numerator and denominator. The derivative of \(5e^{5t}+2t\) is \(25e^{5t}+2\), and the derivative of \(10e^{5t}-1\) is \(50e^{5t}\). So we have \(\lim_{t
ightarrow\infty}\frac{25e^{5t}+2}{50e^{5t}}\).

Step4: Simplify the limit

\(\lim_{t
ightarrow\infty}\frac{25e^{5t}+2}{50e^{5t}}=\lim_{t
ightarrow\infty}(\frac{25e^{5t}}{50e^{5t}}+\frac{2}{50e^{5t}})\). As \(t
ightarrow\infty\), \(\frac{2}{50e^{5t}}
ightarrow0\). And \(\frac{25e^{5t}}{50e^{5t}}=\frac{1}{2}\).

Answer:

\(\frac{1}{2}\)