Question

use lhôpitals rule to evaluate lim 4 sin (7t^7) / 5t as t→0
lim 4 sin (7t^7) / 5t = □ (type an exact answer.)

Explanation:

Step1: Check the form of the limit

When \(t
ightarrow0\), \(\sin(7t^{7})
ightarrow0\) and \(5t
ightarrow0\). So, \(\lim_{t
ightarrow0}\frac{4\sin(7t^{7})}{5t}\) is in the \(\frac{0}{0}\) form.

Step2: Apply L'Hôpital's Rule

Differentiate the numerator and the denominator.
The derivative of \(y = 4\sin(7t^{7})\) using the chain - rule: \(y^\prime=4\cos(7t^{7})\cdot49t^{6}\) (since \(\frac{d}{dt}\sin(u)=\cos(u)\cdot u^\prime\) and \(u = 7t^{7}\), \(u^\prime=49t^{6}\)).
The derivative of \(y = 5t\) is \(y^\prime=5\).
So, \(\lim_{t
ightarrow0}\frac{4\sin(7t^{7})}{5t}=\lim_{t
ightarrow0}\frac{4\cos(7t^{7})\cdot49t^{6}}{5}\).

Step3: Evaluate the new limit

As \(t
ightarrow0\), \(\cos(7t^{7})
ightarrow\cos(0) = 1\) and \(t^{6}
ightarrow0\).
\(\lim_{t
ightarrow0}\frac{4\cos(7t^{7})\cdot49t^{6}}{5}=\frac{4\times1\times0}{5}=0\).

Answer:

\(0\)