We are tasked with evaluating the limit:

$\lim_{x \to 5} \frac{x^2 - 6x + 5}{x^2 - 8x + 15}.$

Step 1: Substitute $x = 5$ into the expression to check for indeterminate forms.

The numerator and denominator are:

• Numerator: $x^2 - 6x + 5 = 5^2 - 6(5) + 5 = 25 - 30 + 5 = 0$,
• Denominator: $x^2 - 8x + 15 = 5^2 - 8(5) + 15 = 25 - 40 + 15 = 0$.

Thus, the expression is of the indeterminate form $\frac{0}{0}$. We must simplify further using algebraic techniques.

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Step 2: Factorize the numerator and denominator.

Factorizing the numerator $x^2 - 6x + 5$:

We seek two numbers whose product is $5$ and sum is $-6$. These numbers are $-5$ and $-1$, so: $x^2 - 6x + 5 = (x - 5)(x - 1).$

Factorizing the denominator $x^2 - 8x + 15$:

We seek two numbers whose product is $15$ and sum is $-8$. These numbers are $-5$ and $-3$, so: $x^2 - 8x + 15 = (x - 5)(x - 3).$

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Step 3: Simplify the fraction.

Substitute the factorizations into the original expression: $\frac{x^2 - 6x + 5}{x^2 - 8x + 15} = \frac{(x - 5)(x - 1)}{(x - 5)(x - 3)}.$

Cancel the common factor $x - 5$ (valid for $x \neq 5$): $\frac{(x - 5)(x - 1)}{(x - 5)(x - 3)} = \frac{x - 1}{x - 3}.$

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Step 4: Evaluate the simplified expression at $x = 5$:

Substitute $x = 5$ into $\frac{x - 1}{x - 3}$: $\frac{x - 1}{x - 3} = \frac{5 - 1}{5 - 3} = \frac{4}{2} = 2.$

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Final Answer:

$\boxed{2}$

Would you like a deeper explanation or have further questions? Here are 5 related questions to expand your understanding:

1. How do we identify and resolve indeterminate forms like $\frac{0}{0}$?
2. Can all rational functions be simplified by factoring?
3. What if the canceled term $x - 5$ had been squared? How would that affect the limit?
4. How do limits work for piecewise functions with discontinuities?
5. What is the relationship between the limit and continuity at a point?

Tip: Always factorize expressions fully when handling limits involving indeterminate forms.