To solve the limit $\lim_{x \to -1} \frac{x+1}{(2x^2 + 7x + 5)^2}$, let's proceed step by step.

Step 1: Plug in $x = -1$

First, substitute $x = -1$ directly into the function:

$f(x) = \frac{x+1}{(2x^2 + 7x + 5)^2}$

Substitute $x = -1$:

$f(-1) = \frac{-1+1}{(2(-1)^2 + 7(-1) + 5)^2} = \frac{0}{(2(1) - 7 + 5)^2}$

$f(-1) = \frac{0}{(2 - 7 + 5)^2} = \frac{0}{0^2} = \frac{0}{0}$

This results in an indeterminate form $\frac{0}{0}$, so we need to apply L'Hôpital's Rule or factor and simplify the expression.

Step 2: Factor the denominator

Let's first factor the quadratic expression in the denominator:

$2x^2 + 7x + 5 = (2x + 5)(x + 1)$

Thus, the original function can be rewritten as:

$f(x) = \frac{x+1}{[(2x + 5)(x + 1)]^2}$

Step 3: Simplify the expression

Notice that the numerator $x + 1$ can cancel out with one $x + 1$ term in the denominator:

$f(x) = \frac{1}{(2x + 5)(x + 1)}$

So, the limit becomes:

$\lim_{x \to -1} \frac{1}{(2x + 5)(x + 1)}$

Step 4: Plug in $x = -1$ again

Substitute $x = -1$ into the simplified expression:

$f(-1) = \frac{1}{(2(-1) + 5)(-1 + 1)} = \frac{1}{( -2 + 5)(0)} = \frac{1}{0}$

However, this gives us an undefined expression because division by zero is undefined. This means that the limit does not exist.

Conclusion

The limit does not exist because the function approaches infinity as $x$ approaches $-1$.

Do you want details or have any questions?

Related Questions:

1. What is L'Hôpital's Rule and when can it be applied?
2. How do you factor quadratic expressions?
3. What are the types of indeterminate forms?
4. How do you analyze limits involving infinity?
5. What does it mean for a limit to not exist?

Tip:

When dealing with indeterminate forms like $\frac{0}{0}$, always consider factoring, simplifying, or applying L'Hôpital's Rule to resolve the limit.