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Chapter 10: Problem 13

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow 0} \frac{2 \cos 2 x-2+4 x^{2}}{2 x^{4}}$$

Short Answer

Expert verified
Answer: The limit of the function as \(x\) approaches 0 is \(\frac{1}{3}\).

Step by step solution

01

Find the Taylor series of \(\cos 2x\) up to and including the \(x^4\) term

The Taylor series of \(\cos 2x\) is given by: $$\cos 2x = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}(2x)^{2n}$$ Now, let's find the first few terms up to and including the \(x^4\) term: $$\cos 2x \approx 1 - \frac{1}{2!}(2x)^2 + \frac{1}{4!}(2x)^4$$ Simplifying the terms, we have: $$\cos 2x \approx 1 - 2x^2 + \frac{1}{3}x^4$$
02

Substitute Taylor series into original expression

Now it's time to plug the Taylor series expansion into our original limit expression: $$\lim _{x \rightarrow 0} \frac{2 \cos 2 x-2+4 x^{2}}{2 x^{4}} = \lim _{x \rightarrow 0} \frac{2(1 - 2x^2 + \frac{1}{3}x^4)-2+4 x^{2}}{2 x^{4}}$$
03

Simplify the expression and evaluate the limit

Let's simplify the expression: $$\lim _{x \rightarrow 0} \frac{2 - 4x^2 + \frac{2}{3}x^4 - 2 + 4 x^{2}}{2 x^{4}} = \lim _{x \rightarrow 0} \frac{\frac{2}{3}x^4}{2 x^{4}}$$ Now we can cancel out the common terms and evaluate the limit: $$\lim _{x \rightarrow 0} \frac{\frac{2}{3}x^4}{2 x^{4}} = \frac{1}{3}$$ So, the limit is: $$\lim _{x \rightarrow 0} \frac{2 \cos 2 x-2+4 x^{2}}{2 x^{4}} = \frac{1}{3}$$

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Most popular questions from this chapter

By comparing the first four terms, show that the Maclaurin series for \(\cos ^{2} x\) can be found (a) by squaring the Maclaurin series for \(\cos x,\) (b) by using the identity \(\cos ^{2} x=(1+\cos 2 x) / 2,\) or \((\mathrm{c})\) by computing the coefficients using the definition.

If the power series \(f(x)=\sum c_{k} x^{k}\) has an interval of convergence of \(|x|

Find the next two terms of the following Taylor series. $$\frac{1}{\sqrt{1+x}} 1-\frac{1}{2} x+\frac{1 \cdot 3}{2 \cdot 4} x^{2}-\frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} x^{3}+\cdots$$

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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. $$(1+x)^{-2}=1-2 x+3 x^{2}-4 x^{3}+\cdots, \text { for }-1 < x < 1$$ $$(1+4 x)^{-2}$$
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