Determine if the function F(x) = (x2 − 6x + 5) / (x2 - 4x + 3) has a removable discontinuity at x=1.

Substitute $x=1$ into the function:

$$F(1) = \frac{1^2 - 6(1) + 5}{1^2 - 4(1) + 3} = \frac{1 - 6 + 5}{1 - 4 + 3} = \frac{0}{0}$$

Since we have an indeterminate form $\frac{0}{0}$, $F(1)$ is undefined.

To evaluate the limit $\lim_{x \to 1} \frac{x^2 - 6x + 5}{x^2 - 4x + 3}$, we factor the numerator and the denominator.
Factor the numerator $x^2 - 6x + 5$:
We look for two numbers that multiply to 5 and add to -6. These are -1 and -5.
So, $x^2 - 6x + 5 = (x-1)(x-5)$.

Factor the denominator $x^2 - 4x + 3$:
We look for two numbers that multiply to 3 and add to -4. These are -1 and -3.
So, $x^2 - 4x + 3 = (x-1)(x-3)$.

Now, substitute the factored forms into the limit expression:

$$\lim_{x \to 1} \frac{(x-1)(x-5)}{(x-1)(x-3)}$$

Since $x \to 1$, $x \neq 1$, so we can cancel the $(x-1)$ terms:

$$\lim_{x \to 1} \frac{x-5}{x-3}$$

Now, substitute $x=1$ into the simplified expression:

$$\frac{1-5}{1-3} = \frac{-4}{-2} = 2$$

The limit $\lim_{x \to 1} F(x)$ exists and is equal to 2.

Since $F(1)$ is undefined but $\lim_{x \to 1} F(x)$ exists (it equals 2), the function $F(x)$ has a removable discontinuity at $x=1$.

The steps shown in your notes are correct.
The function $F(x) = \frac{x^2 - 6x + 5}{x^2 - 4x + 3}$ has a removable discontinuity at $x=1$. The value that would make the function continuous at $x=1$ is $\boxed{2}$.