We are tasked with evaluating the following limit using l'Hôpital's Rule:

$\lim_{{x \to -5}} \frac{x^2 + 14x + 45}{x^2 - 4x - 45}$

Step 1: Check for indeterminate form

Substitute $x = -5$ into the expression:

• For the numerator: $(-5)^2 + 14(-5) + 45 = 25 - 70 + 45 = 0$

• For the denominator: $(-5)^2 - 4(-5) - 45 = 25 + 20 - 45 = 0$

Thus, we have the indeterminate form $\frac{0}{0}$. This indicates that l'Hôpital's Rule can be applied.

Step 2: Apply l'Hôpital's Rule

l'Hôpital's Rule states that if the limit gives an indeterminate form like $\frac{0}{0}$, we can differentiate the numerator and the denominator separately and then take the limit again. Let's compute the derivatives of the numerator and denominator:

• The derivative of the numerator $x^2 + 14x + 45$ is: $2x + 14$

• The derivative of the denominator $x^2 - 4x - 45$ is: $2x - 4$

Thus, applying l'Hôpital's Rule gives us: $\lim_{{x \to -5}} \frac{2x + 14}{2x - 4}$

Step 3: Substitute $x = -5$ again

Substitute $x = -5$ into the simplified limit expression:

• For the numerator: $2(-5) + 14 = -10 + 14 = 4$
• For the denominator: $2(-5) - 4 = -10 - 4 = -14$

So, the limit becomes: $\frac{4}{-14} = -\frac{2}{7}$

Final Answer:

$\lim_{{x \to -5}} \frac{x^2 + 14x + 45}{x^2 - 4x - 45} = -\frac{2}{7}$

Would you like more details or have any questions?

5 Related Questions:

1. How do you determine if l'Hôpital's Rule is applicable?
2. Can l'Hôpital's Rule be used for limits that tend to infinity?
3. What are the conditions under which l'Hôpital's Rule fails?
4. How do you find the derivative of a quotient without using l'Hôpital's Rule?
5. Can indeterminate forms like $\frac{0}{\infty}$ or $\frac{\infty}{\infty}$ be solved using l'Hôpital's Rule?

Tip:

Always check for indeterminate forms before applying l'Hôpital's Rule to ensure its applicability!