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Chapter 10: Problem 18

Evaluate the following limits using Taylor series. $$\lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1-(x / 2)}{4 x^{2}}$$

Short Answer

Expert verified
Question: Determine the limit of the given function as x approaches 0 using Taylor series: $$ \lim _{x \rightarrow 0} \frac{\sqrt{1+x}-1-(x / 2)}{4 x^{2}} $$ Answer: The limit of the given function as x approaches 0 is \(-\frac{1}{32}\).

Step by step solution

01

Expand the Taylor series

Recall the general form of a Taylor series centered at a is: $$ f(x) = \sum_{n=0}^{\infty} \frac{f^{(n)}(a)}{n!}(x-a)^n $$ For our function, we want to find the Taylor series centered at 0 (Maclaurin series) and expand it up to the second-degree term. So, let's find the Maclaurin series for the given function. The function inside the limit is: $$ f(x) = \frac{\sqrt{1+x}-1-(x / 2)}{4 x^{2}} $$ The term inside the square root can be expanded using the binomial theorem letting \(n = 1/2\) and \(a = 0\): $$ \sqrt{1+x} = (1+x)^{1/2} \approx 1 + \frac{1}{2}x - \frac{1}{8}x^2 + \cdots $$ Now, let's substitute the approximation back into the function.
02

Simplify the expression

Replace the square root term with its Taylor series expansion and simplify the expression: $$ f(x) = \frac{(1 + \frac{1}{2}x - \frac{1}{8}x^2 + \cdots) - 1 - (x / 2)}{4 x^{2}} $$ Now remove the common terms and simplify: $$ f(x) = \frac{- \frac{1}{8}x^2}{4 x^{2}} $$
03

Evaluate the limit

Finally, let's evaluate the limit as x approaches 0: $$ \lim _{x \rightarrow 0} \frac{- \frac{1}{8}x^2}{4 x^{2}} =\lim _{x \rightarrow 0} \frac{- 1}{8} \cdot \frac{x^2}{4x^2} =\lim _{x \rightarrow 0} \frac{- 1}{8} \cdot \frac{1}{4} $$ The limit is just the constant term: $$ \lim _{x \rightarrow 0} \frac{- 1}{8} \cdot \frac{1}{4} =\frac{-1}{32} $$ So, the limit of the given function as x approaches 0 is \(-\frac{1}{32}\).

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Use the identity \(\sec x=\frac{1}{\cos x}\) and long division to find the first three terms of the Maclaurin series for \(\sec x\)

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