Question

Question: In general, we do not expect Maclaurin series to be useful in evaluating indeterminate forms except when xtends to zero (see Problem 24). Show, however, that Problem 24(f) can be done by writing and using the series (13.3) for. Hint:Divide numerator and denominator bybefore you take the limit. What is special about the

series which makes it possible to know what the limit of the infinite series is?

Short answer

The solution is

Step-by-step solution

Given information

L’Hopital’s Rule is to be used to evaluate the value of expressions under limits.

 Definition of Maclaurin Series

The definition of the Maclaurin series used in this solution is given.$e^x=1+x+\frac{x^2}{2}+\frac{x^3}{3}+......+\frac{x^n}{n}+\frac{x^{n+1}n}{n+1}+.......$

Begin by applying the Maclaurin series to the given limit. 

Write the given limit.

Write the Maclaurin’s series.

Apply limit to the Maclaurin’s series.

Apply limits to the Maclaurin’s Series.

Multiply and divide the numerator and denominator by.

$$\begin{gathered} \underset{x\to \infty }{\lim }\frac{x^n}{1+x+\frac{x}{2}+\frac{x}{3}+....+\frac{x^n}{n}+\frac{x^{n+1}}{n+1}+.....} \\ =\underset{x\to \infty }{\lim }\frac{1}{\frac{1}{x^n}+\frac{1}{x^{n-1}}+\frac{1}{x^{n-2}2}+\frac{1}{x^{n-3}3}+....{\displaystyle \frac{1}{n}}+...\frac{x}{n+1}+...} \end{gathered}$$

$$\begin{gathered} =\frac{1}{\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{\infty }+\frac{1}{\infty }+....+\frac{1}{n}+\frac{\infty }{n+1}+\infty +...} \\ =\frac{1}{0+\frac{1}{0}+\infty } \end{gathered}$$

Continue the evaluation.

Thus, the solution is.

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