Question

Evaluate the following limits or state that they do not exist. \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x},\) where \(a\) and \(b\) are constants with \(b \neq 0\)

Short answer

$$ \lim_{x \rightarrow 0} \frac{\sin{(ax)}}{\sin{(bx)}} $$ Answer: The limit of the given expression as x approaches 0 is the ratio of the constants \(a\) and \(b\), or \(\frac{a}{b}\).

Step-by-step solution

Verify that the limit is of the form \(\frac{0}{0}\).

As \(x\) approaches 0, both the numerator and the denominator of the given expression approach 0. Specifically, $$ \lim_{x\rightarrow 0}\sin{(ax)} = \lim_{x\rightarrow 0}\sin{(bx)} = 0 $$ since the sine function is continuous and \(\sin{0}=0\). Therefore, the limit is in the form of \(\frac{0}{0}\).

Use L'Hôpital's Rule.

Apply L'Hôpital's Rule, which states that the limit of the ratio of the derivatives is equal to the limit of the original function: $$ \lim_{x \rightarrow 0} \frac{\sin{(ax)}}{\sin{(bx)}} = \lim_{x \rightarrow 0} \frac{d(\sin{(ax)})/dx}{d(\sin{(bx)})/dx} $$

Differentiate both numerator and denominator.

Using the Chain Rule, the derivative of the sine function is calculated as follows: $$ \frac{d(\sin{(ax)})}{dx}=a\cos{(ax)} $$ and $$ \frac{d(\sin{(bx)})}{dx}=b\cos{(bx)} $$ Then plug in these derivatives into the limit: $$ \lim_{x \rightarrow 0} \frac{a\cos{(ax)}}{b\cos{(bx)}} $$

Evaluate the limit.

As \(x\) approaches 0, both the cosine terms in the numerator and the denominator approach 1, and thus the limit becomes: $$ \lim_{x \rightarrow 0} \frac{a\cos{(ax)}}{b\cos{(bx)}} = \lim_{x \rightarrow 0} \frac{a}{b} $$ Since \(a\) and \(b\) are constants, the limit of the expression is simply the ratio \(\frac{a}{b}\).

Key concepts

Limits

Limits are fundamental in calculus, serving as the approach to understanding how functions behave as inputs approach certain points. In the exercise given, we explore the limit of \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\). When evaluating this expression, we see both the numerator, \(\sin(ax)\), and the denominator, \(\sin(bx)\), approach 0 as \(x\) approaches 0. This results in an indeterminate form \(\frac{0}{0}\), which typically requires additional techniques like L'Hôpital's Rule for evaluation.
Indeterminate forms reveal certain ambiguities when trying to evaluate limits directly by substitution, making methods like L'Hôpital's crucial for finding meaningful results.

Trigonometric Functions

Trigonometric functions like sine and cosine play a key role in many calculus problems, especially those involving limits. In our exercise, the functions \(\sin(ax)\) and \(\sin(bx)\) are part of evaluating the limit.
Key properties of sine include its periodicity and range, as well as its derivative. As \(x\) approaches 0, \(\sin(x)\) behaves smoothly, and its value becomes extremely small, similar to its angle in radians. This close approximation to zero is what leads to the indeterminate form scenario. Knowing these properties helps in evaluating expressions around such critical points without hitting computational roadblocks.

Chain Rule

The Chain Rule is an essential tool in calculus for differentiating composite functions. It comes into play in this problem when differentiating functions like \(\sin(ax)\) and \(\sin(bx)\).
According to the Chain Rule, the derivative of \(\sin(ux)\) with respect to \(x\) is \(u \cos(ux)\), where \(u\) is the constant multiplier. By applying the Chain Rule:

• The derivative of \(\sin(ax)\) is \(a\cos(ax)\)
• The derivative of \(\sin(bx)\) is \(b\cos(bx)\)

This frame allows us to substitute these derivatives back into the L'Hôpital's Rule formulation, turning an indeterminate form into an approachable limit that can ultimately calculate the final result: \(\frac{a}{b}\). Understanding the Chain Rule thereby simplifies the process of solving many complex calculus problems involving composite functions.