/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 51 Use series to cvaluate the limit... [FREE SOLUTION] | 91Ó°ÊÓ

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Use series to cvaluate the limit. $$\lim _{x \rightarrow 0} \frac{x-\ln (1+x)}{x^{2}}$$

Short Answer

Expert verified
The limit is \(\frac{1}{2}\).

Step by step solution

01

Recall the Series Expansion for ln(1+x)

The Taylor series expansion for \( ext{ln}(1+x)\) around \(x=0\) is given by:\[\text{ln}(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots\]This series is valid for \(-1 < x \leq 1\).
02

Substitute the Series into the Original Expression

Substitute the series expansion from Step 1 into the given expression:\[x - \text{ln}(1+x) = x - \left( x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \right)\]Simplifying, we get:\[x - \text{ln}(1+x) = \frac{x^2}{2} - \frac{x^3}{3} + \cdots\]
03

Divide the Simplified Expression by x^2

Now, consider the limit expression:\[\frac{x - \text{ln}(1+x)}{x^2} = \frac{\frac{x^2}{2} - \frac{x^3}{3} + \cdots}{x^2}\]Simplify the division by \(x^2\):\[= \frac{1}{2} - \frac{x}{3} + \cdots\]
04

Evaluate the Limit as x Approaches 0

Evaluate the limit:\[\lim_{x \to 0} \left( \frac{1}{2} - \frac{x}{3} + \cdots \right)\]As \(x\) approaches 0, all terms involving \(x\) vanish, leaving us with:\[\frac{1}{2}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Taylor Series
A Taylor series is a way to represent a function as an infinite sum of terms that are derived from the values of its derivatives at a single point. This is an incredibly useful tool in calculus for approximating complex functions using polynomials. The basic idea is to express a function as:\[ f(x) = f(a) + f'(a)(x-a) + \frac{f''(a)}{2!}(x-a)^2 + \frac{f'''(a)}{3!}(x-a)^3 + \cdots \]For the natural logarithm function, this series becomes especially useful for evaluating limits. Specifically, for \( \ln(1+x) \), the Taylor series around \( x = 0 \) is:\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots \]In practice, using only the first few terms often provides a good approximation. This simplification becomes more accurate as \( x \) nears zero, making it ideal for problems involving limit evaluations.
Limit Evaluation
In calculus, evaluating the limit of a function helps us understand its behavior as the input approaches a particular value. When directly substituting this value results in an indefinite form, such as \(\frac{0}{0}\), we must find alternate strategies. Series expansions, like the Taylor series, can simplify this process.
When we substitute the Taylor series expansion of \( \ln(1+x) \) into the expression \( \frac{x - \ln(1+x)}{x^2} \), we simplify to:\[ \frac{\frac{x^2}{2} - \frac{x^3}{3} + \cdots}{x^2} = \frac{1}{2} - \frac{x}{3} + \cdots \]As \( x \) approaches zero, terms involving \( x \) vanish and the limit of the expression is simply the constant term.
Thus, by applying the concept of limits and simplifying with series, the limit evaluates to \( \frac{1}{2} \). This showcases how combining different calculus concepts provides an elegant way to solve complex limits.
Natural Logarithm Series
The natural logarithm function, denoted as \( \ln(x) \), is a fundamental mathematical function often used in calculus. The series expansion for \( \ln(1+x) \) around \( x = 0 \) simplifies the logarithm into a form that is easy to compute with:\[ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots \]This specific series expansion is valid within the interval \(-1 < x \leq 1\). It allows us to break down the logarithm into an infinite polynomial, making it easier to approximate and integrate into other expressions.
In contexts like limit evaluation, the series expansion of the natural logarithm is favorable because it allows terms to cancel or simplify substantially. Each term in the series represents a factor of polynomial growth, helping to predict and assess behavior near \( x = 0 \). For computational purposes, this aids in resolving complex expressions to evaluate limits, like demonstrating \( \lim_{x \to 0} \frac{x - \ln(1+x)}{x^2} = \frac{1}{2} \), by clarifying which terms shrink to zero, resulting in a finite, approachable value.

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Most popular questions from this chapter

After injection of a dose \(D\) of insulin, the concentration of insulin in a patient's system decays exponentially and so it can be written as \(D e^{-a t}\) , where \(t\) represents time in hours and \(a\) is a positive constant. (a) If a dose \(D\) is injected every \(T\) hours, write an expression for the sum of the residual concentrations just before the \((n+1)\) st injection. (b) Determine the limiting pre-injection concentration. (c) If the concentration of insulin must always remain at or above a critical value \(C,\) determine a minimal dosage \(D\) in terms of \(C, a,\) and \(T\)

Use series to approximate the definite integral to within the indicated accuracy. $$\int_{0}^{1} x \cos \left(x^{3}\right) d x \quad(\text { three decimal places })$$

If a water wave with length \(L\) moves with velocity \(v\) across a body of water with depth \(d,\) as in the figure on page \(496,\) then $$v^{2}=\frac{g L}{2 \pi} \tanh \frac{2 \pi d}{L}$$ (a) If the water is deep, show that \(v \approx \sqrt{g L /(2 \pi)}\) (b) If the water is shallow, use the Maclaurin series for tanh to show that \(v \approx \sqrt{g d} .\) (Thus in shallow water the velocity of a wave tends to be independent of the length of the wave.) (c) Use the Alternating Series Estimation Theorem to show that if \(L>10 d\) , then the estimate \(v^{2} \approx g d\) is accurate to within 0.014\(g L .\)

Use a power series to approximate the definite integral to six decimal places. $$ \int_{0}^{0.1} x \arctan (3 x) d x $$

What is wrong with the following calculation? $$\begin{aligned} 0 &=0+0+0+\cdots \\ &=(1-1)+(1-1)+(1-1)+\cdots \\\ &=1-1+1-1+1-1+\cdots \\ &=1+(-1+1)+(-1+1)+(-1+1)+\cdots \\\ &=1+0+0+0+\cdots=1 \end{aligned}$$ (Guido Ubaldus thought that this proved the existence of God because "something has been created out of nothing.")

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