Question

use lhôpitals rule to evaluate $lim_{t \to 0} \frac{5sin(5t^{4})}{-2t}$. $lim_{t \to 0} \frac{5sin(5t^{4})}{-2t}=square$ (type an exact answer.)

Explanation:

Step1: Check the form

As \(t
ightarrow0\), \(\sin(5t^{4})
ightarrow0\), so the limit \(\lim_{t
ightarrow0}\frac{5\sin(5t^{4})}{- 2t}\) is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hopital's Rule

Differentiate the numerator and denominator. The derivative of \(y = 5\sin(5t^{4})\) using the chain - rule: \(y^\prime=5\cos(5t^{4})\cdot(20t^{3}) = 100t^{3}\cos(5t^{4})\), and the derivative of \(y=-2t\) is \(y^\prime=-2\). So the limit becomes \(\lim_{t
ightarrow0}\frac{100t^{3}\cos(5t^{4})}{-2}\).

Step3: Evaluate the new limit

Substitute \(t = 0\) into \(\frac{100t^{3}\cos(5t^{4})}{-2}\). Since \(\cos(0)=1\), we have \(\frac{100\times0^{3}\times\cos(0)}{-2}=0\).

Answer:

\(0\)