Evaluate the algebraic limit: \lim_{x \to 2} \frac{x^3 - 8}{x^2 + x - 6} What is the value… — JAMB UTME

Evaluate the algebraic limit:

\lim_{x \to 2} \frac{x^3 - 8}{x^2 + x - 6}

What is the value of this limit?

The correct value of the limit is

\frac{12}{5}

Evaluating the limit by direct substitution gives the indeterminate form

\frac{0}{0}

. Factoring the numerator

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

and denominator

x^2 + x - 6 = (x - 2)(x + 3)

allows cancellation of

(x - 2)

. Evaluating

\frac{x^2 + 2x + 4}{x + 3}

x = 2

yields

\frac{12}{5}

Check for direct substitution

Substituting

x = 2

gives

\frac{2^3 - 8}{2^2 + 2 - 6} = \frac{0}{0}

, which is an indeterminate form.

Direct substitution results in

\frac{0}{0}

, requiring algebraic factorization.

Factor the numerator and the denominator

Numerator:

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

Denominator:

x^2 + x - 6 = (x - 2)(x + 3)

Use the difference of cubes formula

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

and quadratic factorization.

Cancel the common factor and compute the limit

\lim_{x \to 2} \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)(x + 3)} = \lim_{x \to 2} \frac{x^2 + 2x + 4}{x + 3} = \frac{2^2 + 2(2) + 4}{2 + 3} = \frac{12}{5}

Eliminating the factor

(x - 2)

removes the removable discontinuity at

x = 2

Limits of Indeterminate Forms (0/0) using Factorization

Question