The Limit — Question 9

Question 9 - Solution

We are given: $$\lim_{x \to 4} \frac{x - 4}{\sqrt{x} - 2}$$

First, observe that: $$x \to 4 \Rightarrow \text{numerator } x - 4 \to 0, \quad \text{denominator } \sqrt{x} - 2 \to 0$$

So, this is an indeterminate form $\frac{0}{0}$. We rationalize the denominator to resolve it.

Multiply numerator and denominator by the conjugate of the denominator: $$\frac{x - 4}{\sqrt{x} - 2} \cdot \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{(x - 4)(\sqrt{x} + 2)}{(\sqrt{x} - 2)(\sqrt{x} + 2)} = \frac{(x - 4)(\sqrt{x} + 2)}{x - 4}$$

Now cancel the common factor $x - 4$, assuming $x \neq 4$: $$= \sqrt{x} + 2$$

Now take the limit: $$\lim_{x \to 4} \sqrt{x} + 2 = \sqrt{4} + 2 = 2 + 2 = \boxed{4}$$