use lhospital to determine the following limit (where t is regarded as a constant): lim(ω→4) (cos(ωt)…

Explanation:

Step1: Check indeterminate form

When $\omega\rightarrow4$, the numerator $\cos(\omega t)-\cos(4t)\rightarrow\cos(4t)-\cos(4t) = 0$ and the denominator $16-\omega^{2}\rightarrow16 - 16=0$. So, it is in the $\frac{0}{0}$ form and L'Hopital's rule can be applied.

Step2: Differentiate numerator and denominator

Differentiate the numerator $y_1=\cos(\omega t)-\cos(4t)$ with respect to $\omega$. Using the chain - rule, $\frac{d y_1}{d\omega}=-t\sin(\omega t)$. Differentiate the denominator $y_2 = 16-\omega^{2}$ with respect to $\omega$, $\frac{d y_2}{d\omega}=-2\omega$.

Step3: Find the new limit

Now, find $\lim_{\omega\rightarrow4}\frac{-t\sin(\omega t)}{-2\omega}$. Substitute $\omega = 4$ into the new function. We get $\frac{-t\sin(4t)}{-2\times4}=\frac{t\sin(4t)}{8}$.

Answer:

$\frac{t\sin(4t)}{8}$