Evaluate the limit: \lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x + 2} - 2} What is the numerical … — JAMB UTME

Evaluate the limit:

\lim_{x \to 2} \frac{x^3 - 8}{\sqrt{x + 2} - 2}

What is the numerical value of this limit?

Cevap: 48

The numerical value of the limit is 48.

Evaluating the limit of

\frac{x^3 - 8}{\sqrt{x + 2} - 2}

x \to 2

gives an indeterminate form

\frac{0}{0}

. Factorizing the numerator gives

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

, and rationalizing the denominator by multiplying numerator and denominator by

(\sqrt{x + 2} + 2)

converts the denominator to

x - 2

. Canceling

(x - 2)

leaves

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

. Evaluating at

x = 2

gives

(4 + 4 + 4)(\sqrt{4} + 2) = 12 \times 4 = 48

Identify the limit form via direct substitution

Substituting

x = 2

yields

\frac{0}{0}

Direct evaluation results in an indeterminate form, requiring algebraic manipulation to eliminate the zero factor.

Factorize the numerator using the difference of cubes formula

x^3 - 8 = (x - 2)(x^2 + 2x + 4)

Exposing the factor

(x - 2)

is essential to resolving the zero denominator.

Rationalize the denominator using its algebraic conjugate

Multiply top and bottom by

(\sqrt{x + 2} + 2)

to get denominator

(x + 2) - 4 = x - 2

Applying

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

eliminates the square root from the denominator.

Cancel the common factor and compute the final value

\lim_{x \to 2} (x^2 + 2x + 4)(\sqrt{x + 2} + 2) = (12)(4) = 48.

With

(x - 2)

cancelled for

x \neq 2

, direct substitution now yields a defined real number.

Limits of Indeterminate Forms using Difference of Cubes and Surd Rationalization

Tahmini Süre:2m 30s

Soru