Evaluate the limit: \lim_{x \to \infty} (\sqrt{x^2 + 6x} - x) What is the numerical value … — JAMB UTME

Evaluate the limit:

\lim_{x \to \infty} (\sqrt{x^2 + 6x} - x)

What is the numerical value of this limit?

To evaluate the limit of

\sqrt{x^2 + 6x} - x

x \to \infty

, multiply and divide by its conjugate

\sqrt{x^2 + 6x} + x

. The numerator simplifies to

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

. Dividing both the numerator and denominator by

x

yields

\frac{6}{\sqrt{1 + 6/x} + 1}

. Taking the limit as

x \to \infty

reduces

\frac{6}{x}

0

, resulting in

\frac{6}{\sqrt{1} + 1} = 3

Identify the indeterminate form

Direct evaluation gives

\infty - \infty

, which is an indeterminate form.

Substitution cannot be applied directly when subtracting infinite limits.

Multiply and divide by the algebraic conjugate

\lim_{x \to \infty} \frac{(\sqrt{x^2 + 6x} - x)(\sqrt{x^2 + 6x} + x)}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{(x^2 + 6x) - x^2}{\sqrt{x^2 + 6x} + x} = \lim_{x \to \infty} \frac{6x}{\sqrt{x^2 + 6x} + x}

The identity

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

eliminates the square root in the numerator.

Factor

x

out of the denominator

\lim_{x \to \infty} \frac{6x}{x \left(\sqrt{1 + \frac{6}{x}} + 1\right)} = \lim_{x \to \infty} \frac{6}{\sqrt{1 + \frac{6}{x}} + 1}

Dividing the numerator and denominator by

x

allows evaluation at infinity.

Compute the limit as

x \to \infty

Since

\lim_{x \to \infty} \frac{6}{x} = 0

, the expression becomes

\frac{6}{\sqrt{1 + 0} + 1} = \frac{6}{2} = 3

Terms with

x

in the denominator approach zero as

x

grows arbitrarily large.

Limits at infinity involving radical indeterminate forms of type

\infty - \infty

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