Question

$\lim_{x\to4}\frac{\sqrt{x^{2}+9}-5}{x-4}$ is equal to

• A. $\frac{2}{5}$
• B. $\frac{8}{25}$
• C. 0
• D. $\frac{8}{5}$
• E. $\frac{4}{5}$

Hint:

Calculus Tip: You can also use L'Hôpital's Rule for this $0/0$ limit! Derivative of top: $\frac{1}{2\sqrt{x^2+9}} \cdot 2x = \frac{x}{\sqrt{x^2+9}}$. Derivative of bottom: $1$. Plugging in $x=4$ immediately gives $\frac{4}{\sqrt{25}} = \frac{4}{5}$.

Answer 1

Answer: (E)

Concept:
When direct substitution results in the indeterminate form $0/0$ involving a square root, multiplying the numerator and the denominator by the conjugate of the radical expression rationalizes it, allowing the problematic $(x-a)$ term to be factored out and canceled.

Step 1: Test direct substitution.
Substitute $x = 4$ into the expression: $$\frac{\sqrt{4^2+9} - 5}{4 - 4} = \frac{\sqrt{25} - 5}{0} = \frac{0}{0}$$ This is an indeterminate form, so we must algebraically manipulate the limit.

Step 2: Multiply by the conjugate.
Multiply the top and bottom by the conjugate of the numerator, which is $(\sqrt{x^2+9} + 5)$: $$L = \lim_{x\rightarrow4} \frac{(\sqrt{x^2+9} - 5)(\sqrt{x^2+9} + 5)}{(x - 4)(\sqrt{x^2+9} + 5)}$$

Step 3: Expand the numerator using difference of squares.
Use $(A-B)(A+B) = A^2 - B^2$: $$L = \lim_{x\rightarrow4} \frac{(x^2 + 9) - 25}{(x - 4)(\sqrt{x^2+9} + 5)}$$ $$L = \lim_{x\rightarrow4} \frac{x^2 - 16}{(x - 4)(\sqrt{x^2+9} + 5)}$$

Step 4: Factor and cancel terms.
Factor the difference of squares in the numerator ($x^2 - 16 = (x-4)(x+4)$): $$L = \lim_{x\rightarrow4} \frac{(x - 4)(x + 4)}{(x - 4)(\sqrt{x^2+9} + 5)}$$ Cancel the $(x-4)$ terms to remove the singularity: $$L = \lim_{x\rightarrow4} \frac{x + 4}{\sqrt{x^2+9} + 5}$$

Step 5: Evaluate the simplified limit.
Apply direct substitution with $x = 4$ again: $$L = \frac{4 + 4}{\sqrt{4^2+9} + 5}$$ $$L = \frac{8}{\sqrt{25} + 5} = \frac{8}{5 + 5} = \frac{8}{10}$$ Simplify the fraction: $$L = \frac{4}{5}$$ Hence the correct answer is (E) $\frac{4{5}$}.