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Chapter 1: Problem 50

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 \(\frac{\cos x-1}{2x^2}\) as \(x\) approaches 0 is -1/4.

Step by step solution

01

Graphical Estimation

First, input the function into a graphing utility. As \(x\) approaches 0, observe the y-values of the function. This will give an estimated value of the limit.
02

Tabular Reinforcement

Next, create a table of the values of \(x\) approaching 0 from both sides. For example, take values for \(x\) as 0.1, 0.01, 0.001, -0.1, -0.01, -0.001, and so forth. Substitute these values into the function \(\frac{\cos x-1}{2x^2}\). As \(x\) approaches 0, the value of the function should Echo the estimate from the graphing method.
03

Analytic Calculation

Finally, use analytic methods to find the limit. We start by applying L'Hopital's rule: since this is a 0/0 indeterminate form, and it's a ratio of two functions, we can differentiate both the numerator and denominator and evaluate the new limit:\n \ \(\lim_{x\to0}\frac{(\cos x-1)'}{(2x^2)'}\). The derivative of \(\cos x\) is \(-\sin x\), and the derivative of 1 is 0. The derivative of \(2x^2\) is \(4x\). So, the new limit is: \(\lim_{x\to0}\frac{-\sin x}{4x}\). This is still a 0/0 form, so we can apply l'Hopital's rule again:\ \(\lim_{x\to0}\frac{(-\sin x)'}{(4x)'} = \lim_{x\to0}\frac{-\cos x}{4}\). As \(x\) approaches 0, \(\cos x\) approaches 1, so the limit is \( \frac{-\cos 0}{4} = -\frac{1}{4}\).

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