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Problem 10

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

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.
See the step by step solution

## 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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