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Complete the table by computing \(f(x)\) at the given values of \(x\). Use these results to estimate the indicated limit (if it exists). $$ \begin{array}{l} f(x)=2 x^{2}-1 ; \lim _{x \rightarrow 1} f(x) \\ \hline x \quad 0.9 \quad 0.99 \quad 0.999 \quad 1.001 \quad 1.01 \quad 1.1 \\ \hline f(\boldsymbol{x}) & & & & & & \\ \hline \end{array} $$

Short Answer

Expert verified
Based on the table and the behavior of the function values as \(x\) approaches 1, we can estimate that the limit \(\lim _{x \rightarrow 1} (2x^2 - 1)\) does exist and is approximately equal to 1.
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Step 1: Compute the values of \(f(x)\) at given x-values

First, we need to compute the value of the function \(f(x)=2x^2-1\) for each given value of \(x\). We have the following values of \(x\): 0.9, 0.99, 0.999, 1.001, 1.01, and 1.1.

Step 2: Fill in the table

Now, let's compute \(f(x)\) for each given \(x\) value and fill in the table. \(f(0.9) = 2(0.9)^2 - 1 = 0.62\) \(f(0.99) = 2(0.99)^2 - 1 = 0.9602\) \(f(0.999) = 2(0.999)^2 -1 = 0.996002\) \(f(1.001) = 2(1.001)^2 - 1 = 1.004002\) \(f(1.01) = 2(1.01)^2 - 1 = 1.0402\) \(f(1.1) = 2(1.1)^2 - 1 = 1.42\) Now, our updated table looks like this: $$ \begin{array}{l} f(x)=2 x^{2}-1 ; \lim _{x \rightarrow 1} f(x) \\ \hline x \quad 0.9 \quad 0.99 \quad 0.999 \quad 1.001 \quad 1.01 \quad 1.1 \\ \hline f(\boldsymbol{x}) \quad 0.62 \quad 0.9602 \quad 0.996002 \quad 1.004002 \quad 1.0402 \quad 1.42 \\ \hline \end{array} $$

Step 3: Examine the behavior of \(f(x)\) as \(x\) approaches 1

To estimate the limit, we must look at the behavior of \(f(x)\) as \(x\) approaches 1. We can see that the values of \(f(x)\) are getting closer to the value 1 as \(x\) approaches 1.

Step 4: Estimate the limit

Given the table, we can observe that the values of \(f(x)\) become closer and closer to 1 as \(x\) approaches 1. Therefore, we can estimate that the limit \(\lim _{x \rightarrow 1} f(x)\) does exist, and that it is approximately equal to 1.

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