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Chapter 1: Problem 80

Use a graphing utility to graph the function and estimate the limit. Use a table to reinforce your conclusion. Then find the limit by analytic methods. $$ \lim _{x \rightarrow 0} \frac{\cos x-1}{2 x^{2}} $$

Short Answer

Expert verified
The limit of the function \(f(x)=\frac{\cos x-1}{2x^2}\) as \(x\) approaches 0 is \(-\frac{1}{4}\).

Step by step solution

01

Graphical estimation of the limit

By plotting the function \(y=\frac{\cos(x)-1}{2x^2}\) on a graphing utility, focus on how the function behaves as x approaches 0. The limit value is where the curve of the function heads as x closes in on 0.
02

Estimating the limit using a table

Create a table of values for \(x\) close to 0 (both negatives and positives like -0.1, -0.01, 0.01, 0.1) and compute the corresponding \(y\) values. If the function has a limit at x = 0, your \(y\) values should be getting closer and closer to a specific number as \(x\) approaches 0.
03

Analytical Solution

In calculating the limit of \(f(x) = \frac{\cos x-1}{2x^2}\) as \(x\) approaches 0 analytically, recognize that the function is in the indeterminate form of 0/0 at \(x=0\). Apply L'Hopital's rule which states that the limit of the quotient of two functions as x approaches a particular value is equal to the limit of the quotients of their derivatives. Differentiate \(\cos x-1\) to get \(-\sin x\) and \(2x^2\) to get \(4x\). The limit expression becomes \(\lim_{x \rightarrow 0} \frac{-\sin x}{4x}\). Apply the trigonometric limit \(\lim_{x \rightarrow 0} \frac{\sin x}{x} = 1\), hence the limit is \( -\frac{1}{4}\)

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