Question

3) \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \)

Ask by Patel Ruiz. in the United Kingdom
Jan 23,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The limit is \( \frac{10}{11} \).

Solution

To evaluate the limit: \[ \lim_{x \rightarrow -5} \frac{x^{2} - 25}{(x + 5)(x - 6)} \] **Step 1: Factor the numerator.** Notice that \(x^2 - 25\) is a difference of squares, which can be factored as: \[ x^2 - 25 = (x - 5)(x + 5) \] **Step 2: Simplify the expression.** Substitute the factored form back into the original limit: \[ \frac{(x - 5)(x + 5)}{(x + 5)(x - 6)} = \frac{x - 5}{x - 6} \quad \text{(for } x \neq -5 \text{)} \] **Step 3: Evaluate the limit.** Now, compute the limit by substituting \(x = -5\) into the simplified expression: \[ \frac{-5 - 5}{-5 - 6} = \frac{-10}{-11} = \frac{10}{11} \] **Final Answer:** \[ \boxed{\dfrac{10}{11}} \]

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To find the limit \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \), we first simplify the expression. Notice that \( x^2 - 25 \) can be factored as \( (x - 5)(x + 5) \). So we rewrite the limit: \[ = \lim_{x \rightarrow -5} \frac{(x - 5)(x + 5)}{(x+5)(x-6)} \] Now, we can cancel the \( (x + 5) \) terms from the numerator and denominator, since \( x \rightarrow -5 \) does not make \( x + 5 \) equal to zero in the context of the limit: \[ = \lim_{x \rightarrow -5} \frac{x - 5}{x - 6} \] Now, we can substitute \(-5\) into the simplified limit: \[ = \frac{-5 - 5}{-5 - 6} = \frac{-10}{-11} = \frac{10}{11} \] So, the limit is: \[ \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} = \frac{10}{11} \]

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