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Chapter 10: Problem 24

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow 0} \frac{(1-2 x)^{-1 / 2}-e^{x}}{8 x^{2}}$$

Short Answer

Expert verified
Question: Evaluate the limit $\lim_{x \to 0} \frac{(1-2x)^{-1/2} - e^x}{8x^2}$ using Taylor series. Answer: The limit is $\frac{1}{16}$.

Step by step solution

01

Taylor series expansion of \((1-2x)^{-1/2}\)

We can expand \((1-2x)^{-1/2}\) using binomial theorem: $$(1-2x)^{-1/2} = 1 - \frac{1}{2}(-2x) + \frac{-1/2(-1/2-1)}{2!}(-2x)^2 + \dots$$ $$ = 1 + x + \frac{3}{4}x^2 + \cdots$$
02

Taylor series expansion of \(e^x\)

We can expand \(e^x\) using its well-known Taylor series expansion: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$
03

Substituting expansions in the limit expression

Now, we substitute the Taylor series expansions of \((1-2x)^{-1/2}\) and \(e^x\) in the limit expression: $$\lim_{x\to0} \frac{(1-2x)^{-1/2} - e^x}{8x^2} = \lim_{x\to0} \frac{(1 + x + \frac{3}{4}x^2 + \cdots) - (1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots)}{8x^2}$$
04

Simplifying the expression and evaluating the limit

The expression in the numerator simplifies to: $$\frac{1}{2}x^2 + \frac{1}{4}x^3 + \cdots$$ Now, let's divide the numerator by \(8x^2\) and consider the limit as \(x \to 0\): $$\lim_{x\to0} \frac{\frac{1}{2}x^2 + \frac{1}{4}x^3 + \cdots}{8x^2} = \lim_{x\to0} (\frac{1}{16} + \frac{1}{32}x + \cdots) = \frac{1}{16}$$ Hence, the limit is: $$\lim_{x \to 0} \frac{(1-2x)^{-1/2} - e^x}{8x^2} = \frac{1}{16}$$

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Most popular questions from this chapter

Use properties of power series, substitution, and factoring of constants to find the first four nonzero terms of the Taylor series centered at 0 for the following functions. Use the Taylor series. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$\frac{1}{(3+4 x)^{2}}$$

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