Problem 13 - Essential Calculus

Question

In Exercises \(7-26,\) evaluate the limit, using \(L\) 'Hôpital's Rule if necessary. (In Exercise \(12, n\) is a positive integer.) \(\lim _{x \rightarrow 0} \frac{\sin 2 x}{\sin 3 x}\)

Step-by-Step Solution

Answer

\(\lim _{x \rightarrow 0} \frac{\sin 2x}{\sin 3x} = \frac{2}{3}\)

Step 1: Direct Substitution

The first step is to try direct substitution by plugging the limit value \(x = 0\) into the original function. Doing so, the function becomes \(\frac{\sin 2(0)}{\sin 3(0)}\), which simplifies to \(\frac{0}{0}\), an undefined form.

Step 2: Apply L'Hôpital's Rule

Since the first step yields an undefined form, L'Hôpital's Rule can be applied. This rule means that the original limit equals to the limit of the derivative of the numerator divided by the derivative of the denominator, if the new limit exists. Therefore, the limit is now redefined as \(\lim_{x \rightarrow 0} \frac{2\cos 2x}{3\cos 3x}\)

Step 3: Calculate the new Limit

Now perform substitution on the newly derived function. Substituting \( x = 0 \) gives \(\frac{2\cos 2(0)}{3\cos 3(0)} = \frac{2}{3}\), which is a valid numerical value. Thus, \(\lim _{x \rightarrow 0} \frac{\sin 2x}{\sin 3x} = \frac{2}{3} \). So, the limit of the original function as \( x \) approaches 0 is \(\frac{2}{3}\).

Key Concepts

Limits in Calculus

In calculus, a limit is a fundamental concept that describes the behavior of a function as its argument approaches a particular point, either from the left, the right, or both. It's the value that a function, or sequence, 'approaches' as the input, or index, approaches some value. Limits help us study and understand the properties of functions, especially concerning continuity and discontinuity.

For instance, the exercise at hand involves finding the limit of a function as the variable x approaches 0. Limits can often predict the value of a function at points where it's not explicitly defined by considering the values nearby.

A key application of limits is in defining derivatives and integrals, which are core to the understanding of differential and integral calculus. Limits are also crucial in understanding asymptotic behavior—how functions behave as we approach infinity or negative infinity—and in the analysis of series and sequences.

Direct Substitution Method

The direct substitution method is often the first technique applied when evaluating limits in calculus. This approach involves simply plugging the value that x is approaching directly into the function, assuming the function is continuous at that point.

This method can provide an immediate answer, but it's not always applicable. When direct substitution results in an undefined expression, such as 0/0, ∞/∞, or -∞/-∞, it indicates an indeterminate form. This form doesn't provide a conclusive answer, and additional techniques must be employed to find the limit.

For the given exercise, direct substitution of x = 0 resulted in the indeterminate form 0/0, prompting the need for further investigation, which is where L'Hôpital's Rule comes into play—the rule directly deals with these indeterminate cases by making use of derivatives.

Derivative Application

The application of derivatives is diverse in calculus, one of the critical uses being in the determination of limits, such as those expressed as an indeterminate form. L'Hôpital's rule takes advantage of derivatives to resolve limits that direct substitution cannot handle.

To apply L'Hôpital's Rule, one takes the derivative of the numerator and the derivative of the denominator of the function separately, and then evaluates the limit of the new function. This procedure can be repeated if necessary, until a limit in a determinate form is found or until the application of L'Hôpital's Rule no longer yields progress.

The solution for the textbook exercise involves using derivatives: after applying L'Hôpital's Rule, the derivatives of \(\sin 2x\) and \(\sin 3x\) with respect to x were calculated as \(\cos 2x\cdot 2\) and \(\cos 3x\cdot 3\) respectively. The limit as x approaches 0 can then be found, offering a concrete result for a situation where direct substitution initially failed. This method underscores the power of derivatives in navigating complex calculus problems.