The given function is:

$f(x) = \frac{1 - x}{(x + 1)(x + 3)}$

You are asked to compute:

$\lim_{x \to -1} f(x)$

Let's first substitute $x = -1$ directly into the function to see if the limit can be evaluated directly or if it leads to an indeterminate form:

$f(-1) = \frac{1 - (-1)}{(-1 + 1)((-1) + 3)} = \frac{1 + 1}{(0)(2)} = \frac{2}{0}$

Since the denominator is zero and the numerator is not zero, this suggests a vertical asymptote at $x = -1$. Therefore, the limit likely approaches either positive or negative infinity, depending on the sign of the function around $x = -1$.

Behavior around $x = -1$:

• For $x \to -1^-$ (approaching from the left):
  • The factor $x + 1$ is negative.
  • The factor $x + 3$ is positive.

Thus, the denominator will be negative, and since the numerator is positive ($1 - x > 0$), the function will approach negative infinity as $x \to -1^-$.

• For $x \to -1^+$ (approaching from the right):
  • The factor $x + 1$ is positive.
  • The factor $x + 3$ is positive.

So the denominator will be positive, and the numerator remains positive, leading the function to approach positive infinity as $x \to -1^+$.

Conclusion:

Since the limit does not approach the same value from both sides, the limit does not exist.

Would you like any further details or have any additional questions?

Here are some related questions for practice:

1. How do you determine if a function has a vertical asymptote?
2. What is the significance of one-sided limits in determining the overall limit?
3. How does the behavior of the numerator and denominator affect the sign of the limit?
4. What is the formal definition of a limit not existing?
5. Can you calculate limits using L'Hopital's Rule in cases of indeterminate forms?

Tip: Always check both one-sided limits when the denominator approaches zero to determine if the limit exists or results in infinity.