To evaluate the limit

$\lim_{{x \to -5}} \left( \frac{1}{{x + 5}} + \frac{11}{{(x + 5)(x - 6)}} \right)$

let's proceed step by step.

Step 1: Recognize the form

As $x$ approaches $-5$, both terms in the expression involve $\frac{1}{{x + 5}}$, which suggests that we might encounter an indeterminate form $\frac{1}{0}$ leading to $\infty$. Let's check the behavior more closely:

• The first term, $\frac{1}{{x + 5}}$, clearly tends to $\infty$ or $-\infty$ depending on the direction of $x$ approaching $-5$.
• The second term $\frac{11}{{(x + 5)(x - 6)}}$ also involves $x + 5$, indicating a similar behavior.

Step 2: Factor and Simplify the Expression

We can factor the expression and analyze it:

= \lim_{{x \to -5}} \left( \frac{1 \cdot (x - 6) + 11}{{(x + 5)(x - 6)}} \right)$$ Simplify the numerator: $$1 \cdot (x - 6) + 11 = x - 6 + 11 = x + 5$$ This gives us: $$\lim_{{x \to -5}} \frac{x + 5}{{(x + 5)(x - 6)}}$$ ### Step 3: Cancel Common Factors The $$x + 5$$ in the numerator and denominator cancels out (except where $$x = -5$$): $$\lim_{{x \to -5}} \frac{1}{{x - 6}}$$ ### Step 4: Evaluate the Simplified Limit Now, substitute $$x = -5$$ into the simplified expression: $$\frac{1}{{-5 - 6}} = \frac{1}{{-11}} = -\frac{1}{11}$$ ### Final Answer Thus, the limit is $$\boxed{-\frac{1}{11}}$$ Would you like any further details or have any questions? ### Relative Questions 1. How do you handle limits that result in indeterminate forms? 2. What are the common techniques to simplify expressions before evaluating limits? 3. How can L'Hôpital's rule be applied to limits with indeterminate forms? 4. What happens to limits at infinity for rational functions? 5. How do you evaluate limits involving trigonometric functions? ### Tip Always check if a function can be factored or simplified before directly applying the limit, as this can often help avoid indeterminate forms.