Apply L'Hôpital's rule.
lim
x→6

x + 19
− 5
x − 6

To solve the limit using L'Hôpital's rule, let's first rewrite the expression given:

\[
\lim_{x \to 6} \frac{x + 19 - 5}{x - 6}
\]

Simplifying the numerator:

\[
x + 19 - 5 = x + 14
\]

So, the limit becomes:

\[
\lim_{x \to 6} \frac{x + 14}{x - 6}
\]

### Step 1: Check the form of the limit

Before applying L'Hôpital's rule, we need to ensure the limit is of the indeterminate form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\). Substituting \(x = 6\):

\[
\frac{6 + 14}{6 - 6} = \frac{20}{0}
\]

Since this results in a form of \(\frac{\text{finite}}{0}\), it is not an indeterminate form. Instead, it suggests that the function has a vertical asymptote at \(x = 6\). Thus, L'Hôpital's rule does not apply in this case.

Since the denominator approaches zero while the numerator approaches a finite value, the limit goes to infinity:

\[
\lim_{x \to 6} \frac{x + 14}{x - 6} = \infty
\]

Therefore, the limit does not exist in the finite sense but diverges to infinity.

### Do you have any questions or want further details?

Here are some related questions to explore:

1. What is L'Hôpital's rule and when can it be applied?
2. How do you determine when a function has a vertical asymptote?
3. What other techniques can be used to evaluate limits besides L'Hôpital's rule?
4. Can limits approaching infinity have meaningful interpretations in real-world contexts?
5. What is the difference between limits that go to infinity and limits that do not exist?

**Tip:** Always check the indeterminate form before applying L'Hôpital's rule; not all \(\frac{0}{0}\) forms need this rule if algebraic simplification works better!

Apply L'Hôpital's rule. lim x→6 (x + 19 − 5) / (x − 6)

Learn how to apply L'Hôpital's Rule to evaluate limits. This example problem explores the limit lim x→6 (x + 19 − 5) / (x − 6) and discusses vertical asymptotes. Suitable for students in Grades 11-12.

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