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?

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.