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Chapter 10: Problem 10

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow 0} \frac{\sin 2 x}{x}$$

Short Answer

Expert verified
Answer: The limit of the expression \(\frac{\sin(2x)}{x}\) as x approaches 0 is 2.

Step by step solution

01

Recall the Taylor series expansion for sin(x)

The Taylor series expansion for sin(x) is given by: $$\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots$$
02

Find the Taylor series expansion for sin(2x)

To find the Taylor series expansion for sin(2x), we will replace 'x' in the sin(x) expansion with '2x': $$\sin(2x) = (2x) - \frac{(2x)^3}{3!} + \frac{(2x)^5}{5!} - \frac{(2x)^7}{7!} + \cdots$$ Simplify the above expression: $$\sin(2x) = 2x - \frac{8x^3}{3!} + \frac{32x^5}{5!} - \frac{128x^7}{7!} + \cdots$$
03

Divide the Taylor series for sin(2x) by x

Now, divide the Taylor series of sin(2x) by x: $$\frac{\sin(2x)}{x} = 2 - \frac{8x^2}{3!} + \frac{32x^4}{5!} - \frac{128x^6}{7!} + \cdots$$
04

Apply the limit as x approaches 0

Let's find the limit of the expression as x approaches 0: $$\lim_{x\to0} \frac{\sin(2x)}{x} = \lim_{x\to0} \left(2 - \frac{8x^2}{3!} + \frac{32x^4}{5!} - \frac{128x^6}{7!} + \cdots\right)$$ As x approaches 0, every term with x in it will approach 0, leaving only the term without x, which is 2: $$\lim_{x\to0} \frac{\sin(2x)}{x} = 2$$ Therefore, the limit of the given expression as x approaches 0 is 2.

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