The Limit — Question 4

Question 4

Evaluate the following limit: $$\lim_{x \to 5} \frac{x^2 - 25}{\sqrt{x + 1} - \sqrt{6}}$$

Provide a clear justification for each step.

Question 4 - Solution

We want to evaluate: $$\lim_{x \to 5} \frac{x^2 - 25}{\sqrt{x + 1} - \sqrt{6}}$$

First, try direct substitution: $$\frac{5^2 - 25}{\sqrt{5 + 1} - \sqrt{6}} = \frac{25 - 25}{\sqrt{6} - \sqrt{6}} = \frac{0}{0}$$

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

Factor the numerator: $$x^2 - 25 = (x - 5)(x + 5)$$

So the limit becomes: $$\lim_{x \to 5} \frac{(x - 5)(x + 5)}{\sqrt{x + 1} - \sqrt{6}}$$

The denominator contains square roots, so we multiply by the conjugate: $$\frac{(x - 5)(x + 5)}{\sqrt{x + 1} - \sqrt{6}} \cdot \frac{\sqrt{x + 1} + \sqrt{6}}{\sqrt{x + 1} + \sqrt{6}}$$

This does not change the value of the expression because we are multiplying by $1$.

Now simplify the denominator using the difference of squares formula: $$(a-b)(a+b)=a^2-b^2$$

Here, $$a=\sqrt{x+1} \qquad \text{and} \qquad b=\sqrt{6}$$

Therefore, $$(\sqrt{x + 1} - \sqrt{6})(\sqrt{x + 1} + \sqrt{6}) = (\sqrt{x+1})^2 - (\sqrt{6})^2$$

$$= x+1 - 6$$

$$= x - 5$$

So the expression becomes: $$\lim_{x \to 5} \frac{(x - 5)(x + 5)(\sqrt{x + 1} + \sqrt{6})}{x - 5}$$

Since we are evaluating the limit as $x$ approaches $5$, we may cancel the common factor $x-5$ for $x \neq 5$: $$\lim_{x \to 5} (x + 5)(\sqrt{x + 1} + \sqrt{6})$$

Now substitute $x=5$: $$(5 + 5)(\sqrt{5 + 1} + \sqrt{6})$$

$$= 10(\sqrt{6} + \sqrt{6})$$

$$= 10(2\sqrt{6})$$

$$= 20\sqrt{6}$$

Therefore, $$\boxed{20\sqrt{6}}$$