Question

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

Ask by Valdez Hanson. in Saudi Arabia
Feb 20,2025

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Tutor-Verified Answer

Answer

The limit is \( \frac{10}{11} \) or approximately 0.90909.

Solution

Calculate the limit \( \lim_{x \rightarrow -5} \frac{x^{2}-25}{(x+5)(x-6)} \). Evaluate the limit by following steps: - step0: Evaluate using transformations: \(\lim _{x\rightarrow -5}\left(\frac{x^{2}-25}{\left(x+5\right)\left(x-6\right)}\right)\) - step1: Reduce the fraction: \(\lim _{x\rightarrow -5}\left(\frac{x-5}{x-6}\right)\) - step2: Rewrite the expression: \(\frac{\lim _{x\rightarrow -5}\left(x-5\right)}{\lim _{x\rightarrow -5}\left(x-6\right)}\) - step3: Evaluate: \(\frac{-10}{\lim _{x\rightarrow -5}\left(x-6\right)}\) - step4: Evaluate: \(\frac{-10}{-11}\) - step5: Reduce the fraction: \(\frac{10}{11}\) The limit of the expression \( \frac{x^{2}-25}{(x+5)(x-6)} \) as \( x \) approaches -5 is \( \frac{10}{11} \) or approximately 0.90909.

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The Deep Dive

To evaluate the limit \( \lim_{x \rightarrow -5} \frac{x^{2}-25}{(x+5)(x-6)} \), we start by substituting \( x = -5 \). Direct substitution gives us: \[ \frac{(-5)^{2}-25}{((-5)+5)((-5)-6)} = \frac{25 - 25}{0 \cdot (-11)} = \frac{0}{0}, \] which is indeterminate. To resolve this, we can factor the numerator. The expression \( x^{2}-25 \) can be factored as: \[ x^{2}-25 = (x-5)(x+5). \] Thus, the limit can be rewritten as: \[ \lim_{x \rightarrow -5} \frac{(x-5)(x+5)}{(x+5)(x-6)}. \] We can cancel the \( (x+5) \) term (as long as \( x \neq -5 \)): \[ = \lim_{x \rightarrow -5} \frac{x-5}{x-6}. \] Now we substitute \( x = -5 \): \[ = \frac{-5-5}{-5-6} = \frac{-10}{-11} = \frac{10}{11}. \] Thus, the limit is: \[ \boxed{\frac{10}{11}}. \]

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