Question

Determine the limit of the transcendental function (if it exists). $$ \lim _{t \rightarrow 0} \frac{\sin 3 t}{2 t} $$

Short answer

The limit of the function as t approaches 0 is \( \frac{3}{2} \).

Step-by-step solution

Applying L'Hopital's Rule

Since the limit \( \lim _{t \rightarrow 0} \frac{\sin 3 t}{2 t} \) is in the indeterminate form 0/0, L'Hopital's rule can be used. This rule states that if the limit of a function is in the form of 0/0 or ∞/∞, then the limit of the function is the same as the limit of the ratio of their derivatives. Therefore, calculate the derivative of the numerator and the derivative of the denominator, which gives \( \lim _{t \rightarrow 0} \frac{ \cos 3t \cdot 3}{2} \).

Evaluating the derivative

Next, substitute t=0 into the derivative to get \( \frac{ \cos 0 \cdot 3}{2} \), which simplifies to \( \frac{3}{2} \).

Key concepts

Limits of Transcendental Functions

Transcendental functions such as trigonometric, exponential, and logarithmic functions often appear in calculus problems, specifically when dealing with limits. Evaluating the limit of transcendental functions can be challenging, especially when they result in indeterminate forms like 0/0 or ∞/∞. To tackle these, mathematicians employ specific strategies, one of which is L'Hopital's Rule.

When directly substituting the value into the function does not work due to an indeterminate form, L'Hopital's Rule provides an approach to find the limit by differentiating the numerator and denominator separately. For instance, in the given exercise, the trigonometric function \(\sin 3t/2t\) as \(t\) approaches 0 leads to such an indeterminate form. By applying this rule, we switch our focus from the original function to the behaviour of its derivatives at the point of interest.

Evaluating Derivatives

Derivatives represent the rate at which a function is changing at any given point and are foundational in calculus. In the context of limits, they can help simplify an otherwise complex expression into a more manageable form. Evaluating derivatives requires knowledge of basic differentiation rules, such as the power rule, product rule, chain rule, and the derivatives of basic transcendental functions.

In the exercise provided, after recognizing the indeterminate form, we differentiate both the numerator (trigonometric function) and the denominator (polynomial function) using these rules. For the sine function in the numerator, we use the chain rule: the derivative of \(\sin u\) with respect to \(u\) is \(\cos u\), and then we multiply it by the derivative of \(3t\) with respect to \(t\), which is 3. The derivative of \(2t\) with respect to \(t\) is simply 2. After differentiating, we must evaluate these derivatives at the specific point that was problematic in the original limit.

Indeterminate Forms

In calculus, indeterminate forms arise when evaluating a limit leads to an expression that lacks a clear value, such as 0/0 or ∞/∞. These forms do not mean the limit does not exist; instead, they signal that a different approach is necessary to find the limit. Examples include 0\(\cdot\)∞, ∞ - ∞, 1^∞, 0^0, and ∞^0.

L'Hopital's Rule is particularly powerful for resolving the indeterminate forms 0/0 and ∞/∞. The rule instructs us to take the derivative of the numerator and denominator separately and then reconsider the limit. There may be instances where applying L'Hopital's Rule once is not enough, and it must be applied repeatedly until the indeterminate form is resolved. In our exercise, L'Hopital's Rule simplifies the limit \(\lim _{t \rightarrow 0} \frac{\sin 3 t}{2 t}\) to an expression which, when evaluated at \(t=0\), clearly gives the limit as \(\frac{3}{2}\), illustrating its power to disentangle indeterminate forms.