Evaluate lim x→-3 (1/(x+1) + 1/(x+3))

Steps to Solve

1. Substituting the limit directly

First, we substitute $x = -3$ into the expression:

$$ \frac{1}{-3 + 1} + \frac{1}{-3 + 3} $$

This simplifies to:

$$ \frac{1}{-2} + \frac{1}{0} $$

2. Identifying the issue

We notice that substituting $x = -3$ in the second term results in a division by zero:

$$ \frac{1}{0} $$

which is undefined. Therefore, we need to analyze the limit further by approaching -3 from the left and the right.

3. Analyzing the left-hand limit

We take values slightly less than -3, for instance, let $x = -3 - \epsilon$ where $\epsilon$ is a small positive number.

$$ \frac{1}{(-3 - \epsilon) + 1} + \frac{1}{(-3 - \epsilon) + 3} $$

This becomes:

$$ \frac{1}{-2 - \epsilon} + \frac{1}{0 - \epsilon} $$

As $\epsilon$ approaches 0 from the positive side, the first term approaches $-\frac{1}{2}$ and the second term approaches $-\infty$.

4. Analyzing the right-hand limit

Now let $x = -3 + \epsilon$. We evaluate the limit:

$$ \frac{1}{(-3 + \epsilon) + 1} + \frac{1}{(-3 + \epsilon) + 3} $$

This simplifies to:

$$ \frac{1}{-2 + \epsilon} + \frac{1}{\epsilon} $$

As $\epsilon$ approaches 0 from the positive side, the first term approaches $-\frac{1}{2}$ and the second term approaches $+\infty$.

5. Conclusion about the limit

From our analysis, we find:

• The left-hand limit approaches $-\infty$
• The right-hand limit approaches $+\infty$

Since the left-hand limit and right-hand limit do not equal, we conclude:

$$ \lim_{x \to -3} \left(\frac{1}{x+1} + \frac{1}{x+3}\right) \text{ does not exist.} $$