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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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Limits and Indeterminate Forms

Limits describe the behavior of a function as the input approaches a particular value. When direct substitution results in an indeterminate form like 0/0, special techniques such as l’Hôpital’s rule are needed to evaluate the limit.

l’Hôpital’s Rule

l’Hôpital’s rule states that if a limit yields an indeterminate form 0/0 or ∞/∞, the limit of the ratio of the functions equals the limit of the ratio of their derivatives, provided this latter limit exists.

Derivative of Trigonometric Functions

Understanding the derivatives of sine and cosine functions is essential when applying l’Hôpital’s rule to trigonometric limits. For example, the derivative of sin(kt) with respect to t is k cos(kt), which helps simplify the limit expression.

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