Question
3) \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \)
Ask by Powers Hall. 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 the numerator \( x^2 - 25 \) is a difference of squares and can be factored as:
\[
x^2 - 25 = (x - 5)(x + 5)
\]
**Step 2: Simplify the Expression**
Substitute the factored form back into the limit:
\[
\lim_{x \rightarrow -5} \frac{(x - 5)(x + 5)}{(x + 5)(x - 6)}
\]
Cancel the common factor \( (x + 5) \) in the numerator and denominator (provided \( x \neq -5 \), which is valid since we're taking a limit):
\[
\lim_{x \rightarrow -5} \frac{x - 5}{x - 6}
\]
**Step 3: Substitute \( x = -5 \)**
Now, substitute \( x = -5 \) into the simplified expression:
\[
\frac{-5 - 5}{-5 - 6} = \frac{-10}{-11} = \frac{10}{11}
\]
**Final Answer:**
\[
\frac{10}{11}
\]
Reviewed and approved by the UpStudy tutoring team
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Simplify this solution Mind Expander
To solve the limit \( \lim _{x \rightarrow-5} \frac{x^{2}-25}{(x+5)(x-6)} \), first notice that directly substituting \( x = -5 \) yields an indeterminate form \(\frac{0}{0}\). To resolve this, we can factor the numerator: \[ x^2 - 25 = (x - 5)(x + 5) \] Now, we can rewrite the limit as: \[ \lim_{x \rightarrow -5} \frac{(x - 5)(x + 5)}{(x + 5)(x - 6)} \] We see that \( (x + 5) \) can be canceled from the numerator and denominator, giving us: \[ \lim_{x \rightarrow -5} \frac{x - 5}{x - 6} \] Now, we can substitute \( x = -5 \): \[ \frac{-5 - 5}{-5 - 6} = \frac{-10}{-11} = \frac{10}{11} \] Thus, the limit is: \[ \boxed{\frac{10}{11}} \]
