The problem asks why the function $f(x)$ is not continuous at $x = 1$. To determine this, let's analyze the given function:

$f(x) = \frac{x^2 + x - 2}{x^2 - 5x + 4}$

Step 1: Factorize the numerator and the denominator.

• Numerator: $x^2 + x - 2$ can be factored as $(x - 1)(x + 2)$.
• Denominator: $x^2 - 5x + 4$ can be factored as $(x - 1)(x - 4)$.

So the function can be rewritten as:

$f(x) = \frac{(x - 1)(x + 2)}{(x - 1)(x - 4)}$

Step 2: Simplify the function.

• The factor $(x - 1)$ appears in both the numerator and denominator, so it can be canceled out for all $x \neq 1$.

Thus, for $x \neq 1$, the function simplifies to:

$f(x) = \frac{x + 2}{x - 4}$

Step 3: Evaluate the conditions for continuity at $x = 1$.

For a function to be continuous at $x = 1$, the following three conditions must be satisfied:

1. $\lim_{x \to 1} f(x)$ exists.
2. $f(1)$ is defined.
3. $\lim_{x \to 1} f(x) = f(1)$.

• Limit: To find $\lim_{x \to 1} f(x)$, we use the simplified function $f(x) = \frac{x + 2}{x - 4}$. As $x$ approaches 1: $\lim_{x \to 1} f(x) = \frac{1 + 2}{1 - 4} = \frac{3}{-3} = -1$

• Function value: To find $f(1)$, substitute $x = 1$ into the original function: $f(1) = \frac{1^2 + 1 - 2}{1^2 - 5(1) + 4} = \frac{0}{0}$ The result $\frac{0}{0}$ is undefined, indicating that $f(1)$ does not exist.

Since $f(1)$ does not exist, the function is not continuous at $x = 1$.

Answer:

The correct choice is B: $\lim_{x \to 1} f(x)$ exists, but $f(1)$ does not exist.

Would you like more details or have any questions?

5 Related Questions:

1. How can you determine if a function is continuous at a different point, say $x = 2$?
2. What is the significance of removing a removable discontinuity?
3. How would the analysis change if $x = 4$ was the point of interest?
4. Can you always cancel out factors like $(x - 1)$ in any rational function?
5. What are other common types of discontinuities besides removable discontinuities?

Tip:

Always check if a function value is defined at the point of interest when testing for continuity, especially if there's an indeterminate form like $\frac{0}{0}$.