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Chapter 10: Problem 12

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

Short Answer

Expert verified
Based on the step-by-step solution, provide a short answer to the problem: The value of the limit $$\lim _{x \rightarrow 0} \frac{1+x-e^{x}}{4 x^{2}}$$ is $$-\frac{1}{8}$$.

Step by step solution

01

Find the Taylor series of e^x around x=0

Recall that the Taylor series expansion of a function \(f(x)\) around \(x=a\) is given by: $$f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f^{(3)}(a)}{3!}(x-a)^3 + \cdots$$ For the function \(e^x\), we know that its derivatives are equal to the function itself and at \(x=0\), it equals to 1. Therefore, the Taylor series expansion of the function \(e^x\) around \(x=0\) is: $$e^x = 1 + x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$
02

Substitute the Taylor series in the limit

Now that we have the Taylor series for \(e^x\) around \(x=0\), we can substitute it in the given limit: $$\lim_{x\to0} \frac{1+x-e^x}{4x^2} = \lim_{x\to0} \frac{1+x-(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots)}{4x^2}$$
03

Simplify the expression and eliminate indeterminate forms

We now simplify the expression by eliminating the terms: $$\lim_{x\to0} \frac{-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots}{4x^2}$$ We can factor out \(x^2\) from the numerator to eliminate any indeterminate forms: $$\lim_{x\to0} \frac{-x^2(\frac{1}{2!}+\frac{x}{3!}+\cdots)}{4x^2} = \lim_{x\to0} \frac{-\frac{1}{2!}-\frac{x}{3!}-\cdots}{4}$$
04

Evaluate the limit

Now, the expression is no longer in an indeterminate form and we can directly substitute \(x=0\) to calculate the limit: $$\lim_{x\to0} \frac{-\frac{1}{2!}-\frac{x}{3!}-\cdots}{4} = \frac{-\frac{1}{2!}}{4} = -\frac{1}{8}$$ So, the value of the limit is: $$\lim _{x \rightarrow 0} \frac{1+x-e^{x}}{4 x^{2}} = -\frac{1}{8}$$

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