Apply L'Hôpital's rule.
lim 
x→6 

x + 19
 − 5
x − 6


The limit you're dealing with is:

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

### Step 1: Check if the limit leads to an indeterminate form
First, we substitute \(x = 6\) directly into the expression:

\[
\frac{6 + 19}{6 - 6} = \frac{25}{0}
\]

Since this results in division by zero, but the numerator does not approach zero (i.e., it's a nonzero constant), **L'Hôpital's rule cannot be applied** here. L'Hôpital's rule is only used for limits that result in an indeterminate form such as \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\).

Instead, you can analyze the behavior of the function as \(x \to 6\).

### Step 2: Analyze the limit
- As \(x\) approaches 6, the denominator \(x - 6\) approaches 0.
- However

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

Learn how to analyze the limit lim x→6 (x + 19) / (x − 6). This detailed solution explains why L'Hôpital's rule cannot be applied, and how to interpret the result when the denominator tends to zero. Suitable for students in Grades 11-12 and early college calculus courses.

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