Problem 67 - Calculus Early Transcendentals

Question

Use Taylor series to evaluate the following limits. Express the result in terms of the parameter(s). $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}$$

Step-by-Step Solution

Answer

Answer: The value of the limit \(\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x}\) is \(\frac{a}{b}\).

Step 1: Taylor series expansion for \(\sin(ax)\)

To find the Taylor series expansion for \(\sin(ax)\), we need to recall the Taylor series for \(\sin(x)\) and substitute \(x\) with \(ax\), which is: $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$ So for \(\sin(ax)\), we have: $$\sin(ax) = ax - \frac{(ax)^3}{3!} + \frac{(ax)^5}{5!} - \frac{(ax)^7}{7!} + \cdots$$

Step 2: Taylor series expansion for \(\sin(bx)\)

Similarly, we find the Taylor series expansion for \(\sin(bx)\) by substituting \(x\) with \(bx\) in the Taylor series for \(\sin(x)\): $$\sin(bx) = bx - \frac{(bx)^3}{3!} + \frac{(bx)^5}{5!} - \frac{(bx)^7}{7!} + \cdots$$

Step 3: Divide the two Taylor series

Now we divide the Taylor series of \(\sin(ax)\) by the Taylor series of \(\sin(bx)\). With this division, we get the new series: $$\frac{\sin(ax)}{\sin(bx)} = \frac{ax - \frac{(ax)^3}{3!} + \frac{(ax)^5}{5!} - \frac{(ax)^7}{7!} + \cdots}{bx - \frac{(bx)^3}{3!} + \frac{(bx)^5}{5!} - \frac{(bx)^7}{7!} + \cdots}$$

Step 4: Evaluate the limit

The next step is to evaluate the limit as \(x\) approaches \(0\): $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} = \lim _{x \rightarrow 0}\frac{ax - \frac{(ax)^3}{3!} + \frac{(ax)^5}{5!} - \frac{(ax)^7}{7!} + \cdots}{bx - \frac{(bx)^3}{3!} + \frac{(bx)^5}{5!} - \frac{(bx)^7}{7!} + \cdots}$$

Step 5: Simplify and express the result in terms of the parameters

Observe that each term in the series for \(\frac{\sin(ax)}{\sin(bx)}\) contains a power of \(x\). This means that as \(x\) approaches \(0\), the higher order terms (those with larger powers of \(x\)) will become negligible compared to the lower order terms. Thus, we can approximate the limit by considering only the first terms in the two series (i.e., those with the lowest powers of \(x\)). This gives us: $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} \approx \lim _{x \rightarrow 0}\frac{ax }{bx } = \frac{a}{b}$$ So the result of the limit is: $$\lim _{x \rightarrow 0} \frac{\sin a x}{\sin b x} = \frac{a}{b}$$