Limits Properties — Question 2

Question 2

Suppose that $$\lim_{x \to 3} f(x) = 5 \quad \text{and} \quad \lim_{x \to 3} g(x) = 0.$$ Define $$h(x) = \frac{f(x)^2 - 25}{g(x)}.$$

• Explain why direct substitution does not determine $\displaystyle \lim_{x \to 3} h(x)$.

• Use algebraic manipulation to rewrite $h(x)$.

• State precisely what additional information would be required to evaluate $\displaystyle \lim_{x \to 3} h(x)$.

Question 2 - Solution

We are given: $$\lim_{x \to 3} f(x) = 5, \quad \lim_{x \to 3} g(x) = 0, \quad h(x) = \frac{f(x)^2 - 25}{g(x)}.$$

(a) Indeterminate form

Substituting the known limits gives: $$\lim_{x \to 3} h(x) = \frac{5^2 - 25}{0} = \frac{0}{0},$$ which is an indeterminate form. Therefore, the limit cannot be evaluated by direct substitution.

(b) Algebraic simplification

Factor the numerator: $$f(x)^2 - 25 = (f(x) - 5)(f(x) + 5).$$ Then, $$h(x) = \frac{(f(x) - 5)(f(x) + 5)}{g(x)} = \left( \frac{f(x) - 5}{g(x)} \right)(f(x) + 5).$$

Taking limits where possible, $$\lim_{x \to 3} (f(x) + 5) = 10.$$

(c) Required additional information

The limit $$\lim_{x \to 3} h(x)$$ exists if and only if the limit $$\lim_{x \to 3} \frac{f(x) - 5}{g(x)}$$ exists.

If we let $$L = \lim_{x \to 3} \frac{f(x) - 5}{g(x)},$$ then $$\lim_{x \to 3} h(x) = 10L.$$

Without information about how fast $f(x) \to 5$ relative to $g(x) \to 0$, the value of the limit cannot be determined.

$$\boxed{\text{The limit cannot be evaluated without additional information.}}$$