Questions: Let f(x) = (x^2 - 9) / (x + 3) Complete the table below. Round all table values to four decimal places, if necessary. Enter DNE if a function value is undefined. x -3.1 -3.01 -3.001 -3.0001 -3 -2.9999 -2.999 -2.99 -2.9 f(x) -6.1 -6.01 -6.001 -6.0001 DNE -5.9999 -5.999 -5.99 -5.9 Using the data in the table above, determine the following limit. Enter DNE if a limit fails to exist, except in case of an infinite limit. If an infinite limit exists, enter ∞ or -∞, as appropriate. lim x -> -3^- (x^2 - 9)/(x + 3) = lim x -> -3^+ (x^2 - 9)/(x + 3) = lim x -> -3 (x^2 - 9)/(x + 3) =

Let f(x) = (x^2 - 9) / (x + 3) Complete the table below. Round all table values to four decimal places, if necessary. Enter DNE if a function value is undefined. x -3.1 -3.01 -3.001 -3.0001 -3 -2.9999 -2.999 -2.99 -2.9 f(x) -6.1 -6.01 -6.001 -6.0001 DNE -5.9999 -5.999 -5.99 -5.9 Using the data in the table above, determine the following limit. Enter DNE if a limit fails to exist, except in case of an infinite limit. If an infinite limit exists, enter ∞ or -∞, as appropriate. lim x -> -3^- (x^2 - 9)/(x + 3) = lim x -> -3^+ (x^2 - 9)/(x + 3) = lim x -> -3 (x^2 - 9)/(x + 3) =
Transcript text: Let $f(x)=\frac{x^{2}-9}{x+3}$ Complete the table below. Round all table values to four decimal places, if necessary. Enter DNE if a function value is undefined. \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} \hline $\boldsymbol{x}$ & -3.1 & -3.01 & -3.001 & -3.0001 & -3 & -2.9999 & -2.999 & -2.99 & -2.9 \\ \hline $\boldsymbol{f}(\boldsymbol{x})$ & -6.1 & -6.01 & -6.00 & -6.00 & DNE & -5.99 & -5.99 & -5.99 & -5.9 \\ \hline \end{tabular} Using the data in the table above, determine the following limit. Enter DNE if a limit fails to exist, except in case of an infinite limit. If an infinite limit exists, enter $\infty$ or $-\infty$, as appropriate. \[ \begin{array}{l} \lim _{x \rightarrow-3^{-}} \frac{x^{2}-9}{x+3}=\square \\ \lim _{x \rightarrow-3^{+}} \frac{x^{2}-9}{x+3}=\square \\ \lim _{x \rightarrow-3} \frac{x^{2}-9}{x+3}=\square \end{array} \]
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Solution

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Solution Steps

Step 1: Evaluate the Function at Given Points

We evaluate the function \( f(x) = \frac{x^2 - 9}{x + 3} \) at the specified \( x \) values: \[ \begin{align_} f(-3.1) & = -6.1, \\ f(-3.01) & = -6.01, \\ f(-3.001) & = -6.001, \\ f(-3.0001) & = -6.0001, \\ f(-3) & = \text{DNE}, \\ f(-2.9999) & = -5.9999, \\ f(-2.999) & = -5.999, \\ f(-2.99) & = -5.99, \\ f(-2.9) & = -5.9. \end{align_} \]

Step 2: Determine the Left-Hand Limit

We find the limit as \( x \) approaches \(-3\) from the left: \[ \lim_{x \to -3^{-}} f(x) = -6.0001. \]

Step 3: Determine the Right-Hand Limit

We find the limit as \( x \) approaches \(-3\) from the right: \[ \lim_{x \to -3^{+}} f(x) = -5.9999. \]

Step 4: Determine the Overall Limit

Since the left-hand limit and right-hand limit are not equal, we conclude that the overall limit does not exist: \[ \lim_{x \to -3} f(x) = \text{DNE}. \]

Final Answer

\[ \lim _{x \rightarrow-3^{-}} \frac{x^{2}-9}{x+3}=\boxed{-6} \\ \lim _{x \rightarrow-3^{+}} \frac{x^{2}-9}{x+3}=\boxed{-6} \\ \lim _{x \rightarrow-3} \frac{x^{2}-9}{x+3}=\boxed{\text{DNE}} \]

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