Use l’Hôpital’s rule to find the limits in Exercises 7–52.

14. lim (t → 0) sin 5t / 2t

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Identify the limit expression: $\lim_{t \to 0} \frac{\sin 5t}{2t}$.

Check if the limit is an indeterminate form by substituting $t = 0$: both numerator $\sin 5(0) = 0$ and denominator $2(0) = 0$, so the form is $\frac{0}{0}$, which allows the use of l'Hôpital's Rule.

Apply l'Hôpital's Rule by differentiating the numerator and denominator separately with respect to $t$: the derivative of the numerator $\sin 5t$ is $5 \cos 5t$, and the derivative of the denominator \$2t$ is $2$.

Rewrite the limit using these derivatives: $\lim_{t \to 0} \frac{5 \cos 5t}{2}$.

Evaluate the new limit by substituting $t = 0$: calculate $\frac{5 \cos 0}{2}$, noting that $\cos 0 = 1$.

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