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The problem provided asks you to evaluate the limit:

\[
\lim_{{x \to 7}} \frac{{x^2 - 7x}}{{x^2 - 4x - 21}}
\]

### Step-by-step solution:

1. **Factor the numerator and denominator:**

   - The numerator \(x^2 - 7x\) can be factored as:
     \[
     x(x - 7)
     \]

   - The denominator \(x^2 - 4x - 21\) is a quadratic expression, so we need to factor it. Find two numbers that multiply to \(-21\) and add to \(-4\). These numbers are \( -7 \) and \( 3 \):
     \[
     x^2 - 4x - 21 = (x - 7)(x + 3)
     \]

2. **Rewrite the limit expression:**

   Substituting the factored forms into the expression:
   \[
   \lim_{{x \to 7}} \frac{{x(x - 7)}}{{(x - 7)(x + 3)}}
   \]

3. **Cancel common terms:**

   The \((x - 7)\) term appears in both the numerator and the denominator, so it can be canceled out:
   \[
   \lim_{{x \to 7}} \frac{x}{{x + 3}}
   \]

4. **Evaluate the limit:**

   Now, substitute \(x = 7\) into the simplified expression:
   \[
   \frac{7}{{7 + 3}} = \frac{7}{10}
   \]

Thus, the value of the limit is:
\[
\boxed{\frac{7}{10}}
\]

Would you like more details or have any questions?

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Here are 5 related questions to practice:

1. How do you factor a quadratic expression like \(x^2 - 4x - 21\)?
2. What is the limit when the numerator and denominator both contain a removable discontinuity?
3. How can you use L'Hopital's Rule if the limit results in a \(\frac{0}{0}\) form?
4. What are some common mistakes to avoid when factoring polynomials?
5. How would you evaluate the limit if the function involved \(x^3 - 27\) in the denominator?

**Tip:** Always check for factorizations in limit problems to simplify expressions before substituting values.

Evaluate the following limit by factoring and cancelling: lim_{x->7} (x^2 - 7x) / (x^2 - 4x - 21).

Learn how to solve limit problems by factoring and cancelling with a detailed example of the limit as x approaches 7 for the expression (x^2 - 7x) / (x^2 - 4x - 21). This tutorial is suitable for high school students studying calculus.

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