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

Short Answer

Expert verified
Based on the step-by-step solution provided, the limit of the given expression as \(x\) approaches 0 is 0. We used the Taylor series of the exponential function \(e^x\) to solve this problem.

Step by step solution

01

Finding the Taylor series for the exponential function

Recall that the Taylor series of the exponential function \(e^x\) is given by: $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$$
02

Substituting the Taylor series into the expression

We will now substitute the Taylor series expansion of \(e^x\) into the expression inside the limit: $$\frac{1 + x - e^x}{4x^2} = \frac{1+x - (1 + x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots)}{4x^2}$$
03

Simplifying the expression

After substituting, let's simplify the expression: $$\frac{1+x - (1 + x + \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots)}{4x^2} = \frac{-\frac{x^2}{2!}-\frac{x^3}{3!}-\cdots}{4x^2} = -\frac{x^2}{8} - \frac{x^3}{24} - \cdots$$
04

Evaluating the limit

Finally, we will evaluate the limit as x approaches 0: $$\lim_{x \rightarrow 0} \left(-\frac{x^2}{8} - \frac{x^3}{24} - \cdots\right) = -\frac{0}{8} -\frac{0}{24} - \cdots = 0$$ So, the limit of the given expression as \(x\) approaches 0 is 0.

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Key Concepts

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

Limits
A limit in mathematics helps us understand how a function behaves as a certain point is approached. The concept of limits is fundamental in calculus, as it allows us to examine function behavior around specific values.
When we talk about \( \lim_{x \rightarrow a} f(x) \), we are essentially asking: what value does \( f(x) \) get closer to as \( x \) gets closer to \( a \)?In the given exercise, we are asked to find the limit of a function as \( x \) approaches 0. The function is \( \frac{1+x-e^x}{4x^2} \).Here's a look at why limits hold such importance:
  • They help in finding derivatives, which show us the rate of change.
  • Limits allow for precise definitions of continuity.
  • They are essential in defining definite integrals.
Evaluating the limit involves determining how the function behaves near 0. Taylor series and other methods can simplify this process. By inserting the Taylor expansion, we can find the limit by examining terms that remain pronounced as \( x \) becomes very small, leading us to the conclusion that the given limit results in 0.
Exponential Function
An exponential function, often expressed as \( e^x \), is a mathematical function capable of modeling constant growth or decay. The symbol \( e \) represents Euler's number, approximately 2.71828, and is crucial in various mathematical and applied contexts.Exponential functions are characterized by their rapid increase or decrease pattern. Some essential properties and applications include:
  • The derivative of \( e^x \) is itself, \( e^x \).
  • It models phenomena like population growth, radioactive decay, and compound interest.
  • Its inverse is the natural logarithm \( \ln x \).
For the exercise involving limits, understanding the exponential function's behavior through a series expansion allows us to simplify complex expressions. By using the Taylor series expansion of \( e^x \), we can break down the function into a sum of polynomials which are easier to manipulate and evaluate within limits, especially as \( x \) tends towards zero.
Series Expansion
Series expansion is a powerful mathematical tool used to represent functions as a sum of terms calculated from that function's derivatives at a certain point. The Taylor series is a specific type of series expansion around a point \( a \). For example, the Taylor series for \( e^x \) is an infinite series:\[ e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots \]Series expansions help us approximate complex functions by polynomials for ease of computation. The main benefits include:
  • Facilitating calculations of limits and integrals.
  • Approximating difficult-to-calculate functions.
  • Gaining insights into the function's behavior near a specific point.
In the given problem, the expansion allows representing \( e^x \) in a polynomial form to substitute into the limit expression \( \frac{1+x-e^x}{4x^2} \). By applying the Taylor series, each polynomial term can be simplified, allowing us to evaluate the limit correctly as \( x \) approaches zero.

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Most popular questions from this chapter

Symmetry a. Use infinite series to show that \(\cos x\) is an even function. That is, show \(\cos (-x)=\cos x\) b. Use infinite series to show that \(\sin x\) is an odd function. That is, show \(\sin (-x)=-\sin x\)

Use the remainder term to find a bound on the error in the following approximations on the given interval. Error bounds are not unique. $$\ln (1+x)=x-\frac{x^{2}}{2} \text {on } [-0.2,0.2]$$

Matching functions with polynomials Match functions a-f with Taylor polynomials \(A-F\) (all centered at 0 ). Give reasons for your choices. a. \(\sqrt{1+2 x}\) b. \(\frac{1}{\sqrt{1+2 x}}\) c. \(e^{2 x}\) d. \(\frac{1}{1+2 x}\) e. \(\frac{1}{(1+2 x)^{3}}\) f. \(e^{-2 x}\) A. \(p_{2}(x)=1+2 x+2 x^{2}\) B. \(p_{2}(x)=1-6 x+24 x^{2}\) C. \(p_{2}(x)=1+x-\frac{x^{2}}{2}\) D. \(p_{2}(x)=1-2 x+4 x^{2}\) E. \(p_{2}(x)=1-x+\frac{3}{2} x^{2}\) F. \(p_{2}(x)=1-2 x+2 x^{2}\)

Approximating \(\ln x\) Let \(f(x)=\ln x,\) and let \(p_{n}\) and \(q_{n}\) be the nth-order Taylor polynomials for \(f\) centered at 1 and \(e\), respectively. a. Find \(p_{3}\) and \(q_{3}\). b. Graph \(f, p_{3},\) and \(q_{3}\) on the interval (0,4]. c. Complete the following table showing the errors in the approximations given by \(p_{3}\) and \(q_{3}\) at selected points. $$\begin{array}{|c|c|c|} \hline x & \left|\ln x-p_{3}(x)\right| & \left|\ln x-q_{3}(x)\right| \\ \hline 0.5 & & \\ \hline 1.0 & & \\ \hline 1.5 & & \\ \hline 2 & & \\ \hline 2.5 & & \\ \hline 3 & & \\ \hline 3.5 & & \\ \hline \end{array}$$ d. At which points in the table is \(p_{3}\) a better approximation to \(f\) than \(q_{3}\) ? Explain your observations.

Fuliptic integrals The period of an undamped pendulum is given by $$ T=4 \sqrt{\frac{\ell}{g}} \int_{0}^{\pi / 2} \frac{d \theta}{\sqrt{1-k^{2} \sin ^{2} \theta}}=4 \sqrt{\frac{\ell}{g}} F(k) $$ where \(\ell\) is the length of the pendulum, \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity, \(k=\sin \frac{\theta_{0}}{2},\) and \(\theta_{0}\) is the initial angular displacement of the pendulum (in radians). The integral in this formula \(F(k)\) is called an elliptic integral, and it cannot be evaluated analytically. Approximate \(F(0.1)\) by expanding the integrand in a Taylor (binomial) series and integrating term by term.

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