What is the numerical value of the limit \lim_{x \to 0} \frac{\sqrt{4 + x} - \sqrt{4 - x}}… — JAMB UTME

What is the numerical value of the limit

\lim_{x \to 0} \frac{\sqrt{4 + x} - \sqrt{4 - x}}{x}

$\frac{1}{2}$ Answer
B
0
C
2
D
1

Answer

The numerical value of the limit is $\frac{1}{2}$.

Rationalizing the numerator by multiplying with its conjugate gives $\frac{2x}{x(\sqrt{4+x} + \sqrt{4-x})}$. Canceling $x$ yields $\frac{2}{\sqrt{4+x} + \sqrt{4-x}}$, which evaluates to $\frac{2}{4} = \frac{1}{2}$ as $x \to 0$.

Step-by-Step Solution

Identify the indeterminate form

Substituting $x = 0$ directly yields $\frac{\sqrt{4} - \sqrt{4}}{0} = \frac{0}{0}$, which is an indeterminate form requiring rationalization.

Direct evaluation cannot give the true limit value when an indeterminate form is encountered.

Rationalize the numerator

Multiply the numerator and denominator by the conjugate $\sqrt{4 + x} + \sqrt{4 - x}$:

\frac{(\sqrt{4 + x} - \sqrt{4 - x})(\sqrt{4 + x} + \sqrt{4 - x})}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{(4 + x) - (4 - x)}{x(\sqrt{4 + x} + \sqrt{4 - x})}

The difference of squares identity $(a - b)(a + b) = a^2 - b^2$ eliminates the radical signs in the numerator.

Simplify the numerator and cancel common terms

$\frac{4 + x - 4 + x}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{2x}{x(\sqrt{4 + x} + \sqrt{4 - x})} = \frac{2}{\sqrt{4 + x} + \sqrt{4 - x}}$

Canceling the common factor $x$ eliminates the zero-causing term in the denominator.

Evaluate the simplified limit as $x \to 0$

$\frac{2}{\sqrt{4 + 0} + \sqrt{4 - 0}} = \frac{2}{2 + 2} = \frac{2}{4} = \frac{1}{2}$

Direct substitution is now valid since the expression is continuous at $x = 0$.

Key Concept

Evaluation of algebraic limits of indeterminate form 0/0 using radical rationalization.

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