Questions: Find the limit L (if it exists). (If the limit is infinite, enter '∞' or '-', as appropriate. If the limit does not otherwise exist, enter DNE.) lim(x -> 3^-) (x-3)/(x^2-9) L=

Find the limit L (if it exists). (If the limit is infinite, enter '∞' or '-', as appropriate. If the limit does not otherwise exist, enter DNE.)

lim(x -> 3^-) (x-3)/(x^2-9)

L=

Transcript text: Find the limit $L$ (if it exists). (If the limit is infinite, enter ' $\infty$ ' or '- ${ }^{\prime}$ ', as appropriate. If the limit does not otherwise exist, enter DNE.) \[ \begin{array}{l} \lim _{x \rightarrow 3^{-}} \frac{x-3}{x^{2}-9} \\ L=\square \end{array} \]

Solution

Solution Steps

To find the limit of the given function as $ x $ approaches 3 from the left, we first simplify the expression. The denominator can be factored as a difference of squares. After simplification, we evaluate the limit by substituting values approaching 3 from the left into the simplified expression.

Step 1: Simplifying the Function

We start with the limit expression: \[ \lim_{x \rightarrow 3^{-}} \frac{x - 3}{x^2 - 9} \] The denominator can be factored as: \[ x^2 - 9 = (x - 3)(x + 3) \] Thus, we can rewrite the function as: \[ \frac{x - 3}{(x - 3)(x + 3)} = \frac{1}{x + 3} \quad \text{for } x \neq 3 \]

Step 2: Evaluating the Limit

Now, we evaluate the limit as $ x $ approaches 3 from the left: \[ \lim_{x \rightarrow 3^{-}} \frac{1}{x + 3} \] Substituting $ x = 3 $: \[ \frac{1}{3 + 3} = \frac{1}{6} \]

Final Answer

The limit $ L $ is: \[ \boxed{\frac{1}{6}} \]

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