Question

use lhôpitals rule to evaluate $limlimits_{t \to 0}\frac{2sin(4t^{7})}{5t}$.
$limlimits_{t \to 0}\frac{2sin(4t^{7})}{5t}=square$ (type an exact answer.)

Explanation:

Step1: Check the form

When \(t
ightarrow0\), \(\sin(4t^{7})
ightarrow0\) and \(5t
ightarrow0\). So, it is in the \(\frac{0}{0}\) form. Apply L'Hôpital's Rule.
Differentiate the numerator and denominator.
The derivative of \(y = 2\sin(4t^{7})\) using the chain - rule \((\sin u)^\prime=\cos u\cdot u^\prime\) (where \(u = 4t^{7}\), \(u^\prime=28t^{6}\)) is \(y^\prime=2\cos(4t^{7})\cdot28t^{6}=56t^{6}\cos(4t^{7})\).
The derivative of \(y = 5t\) is \(y^\prime = 5\).
So, \(\lim_{t
ightarrow0}\frac{2\sin(4t^{7})}{5t}=\lim_{t
ightarrow0}\frac{56t^{6}\cos(4t^{7})}{5}\).

Step2: Evaluate the limit

As \(t
ightarrow0\), \(\cos(4t^{7})
ightarrow\cos(0) = 1\) and \(t^{6}
ightarrow0\).
\(\lim_{t
ightarrow0}\frac{56t^{6}\cos(4t^{7})}{5}=\frac{56\times0\times1}{5}\).

Answer:

\(0\)