We are tasked with evaluating the limit:

$\lim_{{x \to -7}} \frac{2x^3 + 11x^2 - 21x}{x^2 + 7x}$

Step 1: Simplify the expression

First, factor both the numerator and the denominator.

• Denominator: $x^2 + 7x$ can be factored as: $x^2 + 7x = x(x + 7)$

• Numerator: $2x^3 + 11x^2 - 21x$ has a common factor of $x$, so we factor out $x$: $2x^3 + 11x^2 - 21x = x(2x^2 + 11x - 21)$ Now, we factor $2x^2 + 11x - 21$. Using the quadratic formula or factoring by grouping, we find: $2x^2 + 11x - 21 = (2x - 3)(x + 7)$ Thus, the numerator becomes: $x(2x - 3)(x + 7)$

Now the expression becomes: $\frac{x(2x - 3)(x + 7)}{x(x + 7)}$

Step 2: Cancel common factors

We can cancel the common factors $x$ and $(x + 7)$, provided $x \neq -7$ (since division by zero is undefined):

\frac{(2x - 3)(x + 7)}{(x + 7)} = 2x - 3 \quad \text{(for x \neq -7)}

Step 3: Evaluate the limit

Now that the expression is simplified to $2x - 3$, we substitute $x = -7$ into this expression:

$2(-7) - 3 = -14 - 3 = -17$

Thus, the limit is:

$\lim_{{x \to -7}} \frac{2x^3 + 11x^2 - 21x}{x^2 + 7x} = -17$

Conclusion:

The value of the limit is $-17$.

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Would you like more details or further explanation?

Here are 5 related questions:

1. How do you factor a cubic polynomial like $2x^3 + 11x^2 - 21x$?
2. What are some common techniques for factoring quadratic expressions?
3. Why is it necessary to cancel common factors when simplifying limits?
4. How do we handle limits that lead to indeterminate forms like $\frac{0}{0}$?
5. Can L'Hopital's Rule be applied to limits involving rational functions?

Tip: Always check for common factors in both the numerator and denominator to simplify complex expressions before evaluating limits!