Evaluate the algebraic limit: \lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 16} What is the exa… — JAMB UTME

Evaluate the algebraic limit:

\lim_{x \to 4} \frac{\sqrt{x} - 2}{x^2 - 16}

What is the exact value of this limit?

The exact value of the limit is

\frac{1}{32}

The limit presents an indeterminate form

\frac{0}{0}

x = 4

. Factoring

x^2 - 16

into

(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)

allows the factor

(\sqrt{x} - 2)

to be cancelled from both the numerator and denominator. Substituting

x = 4

into the simplified expression

\frac{1}{(\sqrt{x} + 2)(x + 4)}

yields

\frac{1}{(2+2)(4+4)} = \frac{1}{32}

Identify the form of the limit by direct substitution.

Substituting

x = 4

into

\frac{\sqrt{4} - 2}{4^2 - 16}

gives

\frac{0}{0}

, which is an indeterminate form.

Direct substitution yields

\frac{0}{0}

, requiring algebraic simplification.

Factor the denominator

x^2 - 16

x^2 - 16 = (x - 4)(x + 4)

Use the difference of two squares identity

a^2 - b^2 = (a - b)(a + b)

Factor

(x - 4)

as a difference of squares involving square roots.

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

This exposes the vanishing factor

(\sqrt{x} - 2)

in the denominator.

Cancel the common factor

(\sqrt{x} - 2)

and evaluate the limit.

\lim_{x \to 4} \frac{\sqrt{x} - 2}{(\sqrt{x} - 2)(\sqrt{x} + 2)(x + 4)} = \lim_{x \to 4} \frac{1}{(\sqrt{x} + 2)(x + 4)} = \frac{1}{(\sqrt{4} + 2)(4 + 4)} = \frac{1}{4 \times 8} = \frac{1}{32}

Cancelling the factor removing the

\frac{0}{0}

condition allows direct evaluation.

Resolution of indeterminate limits of the form 0/0 using algebraic factorization and conjugate radical identities.

Tahmini Süre:2m 0s

Soru