Question

The value of $\lim_{x\rightarrow 0}\frac{\sqrt{1+3x}-\sqrt{1-3x}}{x}$ is

• A. 0
• B. 1
• C. $\frac{1}{3}$
• D. $\frac{1}{2}$
• E. 3

Hint:

Logic Tip: You can also use L'Hôpital's rule. The derivative of $\sqrt{1 \pm 3x}$ is $\frac{\pm 3}{2\sqrt{1 \pm 3x}}$. Applying this to the numerator gives $\frac{3}{2(1)} - \frac{-3}{2(1)} = \frac{3}{2} + \frac{3}{2} = 3$. The denominator derivative is 1. This provides the answer immediately!

Answer 1

Answer: (E)

Concept:
Direct substitution of $x=0$ yields the indeterminate form $\frac{0}{0}$. To evaluate limits involving square roots that result in $\frac{0}{0}$, standard practice is to multiply the numerator and the denominator by the conjugate of the radical expression to eliminate the roots.
Step 1: Multiply by the conjugate of the numerator.
The conjugate of $\sqrt{1+3x} - \sqrt{1-3x}$ is $\sqrt{1+3x} + \sqrt{1-3x}$. $$\lim_{x\rightarrow 0} \left[ \frac{\sqrt{1+3x}-\sqrt{1-3x}}{x} \cdot \frac{\sqrt{1+3x}+\sqrt{1-3x}}{\sqrt{1+3x}+\sqrt{1-3x}} \right]$$
Step 2: Simplify the numerator using difference of squares.
The numerator becomes $(a-b)(a+b) = a^2 - b^2$: $$= \lim_{x\rightarrow 0} \frac{(\sqrt{1+3x})^2 - (\sqrt{1-3x})^2}{x(\sqrt{1+3x}+\sqrt{1-3x})}$$ $$= \lim_{x\rightarrow 0} \frac{(1+3x) - (1-3x)}{x(\sqrt{1+3x}+\sqrt{1-3x})}$$ $$= \lim_{x\rightarrow 0} \frac{1 + 3x - 1 + 3x}{x(\sqrt{1+3x}+\sqrt{1-3x})}$$ $$= \lim_{x\rightarrow 0} \frac{6x}{x(\sqrt{1+3x}+\sqrt{1-3x})}$$
Step 3: Cancel out the common x term.
Since $x \rightarrow 0$ but $x \neq 0$, we can cancel $x$ from the numerator and denominator: $$= \lim_{x\rightarrow 0} \frac{6}{\sqrt{1+3x}+\sqrt{1-3x}}$$
Step 4: Apply direct substitution.
Now substitute $x = 0$ into the simplified limit: $$= \frac{6}{\sqrt{1+0} + \sqrt{1-0}}$$ $$= \frac{6}{\sqrt{1} + \sqrt{1}}$$ $$= \frac{6}{1 + 1} = \frac{6}{2} = 3$$