The Limit — Question 6

Question 6 - Solution

We want to evaluate: $$\lim_{x \to 2} \frac{x^2 - 4}{x^3 - 8}$$

First, try direct substitution: $$\frac{2^2 - 4}{2^3 - 8} = \frac{4 - 4}{8 - 8} = \frac{0}{0}$$

This is an indeterminate form, so we need to simplify the expression before evaluating the limit.

Now factor the numerator: $$x^2 - 4 = (x - 2)(x + 2)$$

This is a difference of squares because: $$x^2 - 4 = x^2 - 2^2$$

Next, factor the denominator: $$x^3 - 8 = x^3 - 2^3$$

This is a difference of cubes. Recall the formula: $$a^3 - b^3 = (a-b)(a^2 + ab + b^2)$$

Here, $$a=x \qquad \text{and} \qquad b=2$$

So: $$x^3 - 8 = (x-2)(x^2 + 2x + 4)$$

Now rewrite the original limit using these factorizations: $$\lim_{x \to 2} \frac{(x - 2)(x + 2)}{(x - 2)(x^2 + 2x + 4)}$$

Both the numerator and denominator contain the factor $x-2$. Since we are finding the limit as $x$ approaches $2$, we only need to consider values of $x$ near $2$, but not equal to $2$. Therefore, for $x \neq 2$, we may cancel the common factor $x-2$:

$$\lim_{x \to 2} \frac{x + 2}{x^2 + 2x + 4}$$

Now the expression is no longer indeterminate, so we can substitute $x=2$:

$$\frac{2 + 2}{2^2 + 2(2) + 4}$$

$$= \frac{4}{4 + 4 + 4}$$

$$= \frac{4}{12}$$

$$= \frac{1}{3}$$

Therefore, $$\boxed{\frac{1}{3}}$$