mathop {lim }limits_{x to infty } (sqrt {{x^2 | vedclass

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Question

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} )$,we rationalize the expression:$\

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(C) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} )$,we rationalize the expression:$\mathop {\lim }\limits_{x 	o \infty } \frac{(\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3})(\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3})}{\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{({x^2} + 8x + 3) - ({x^2} + 4x + 3)}{\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{4x}{\sqrt {{x^2} + 8x + 3} + \sqrt {{x^2} + 4x + 3}}$Divide the numerator and denominator by $x$ (for $x > 0$):$= \mathop {\lim }\limits_{x 	o \infty } \frac{4}{\sqrt {1 + \frac{8}{x} + \frac{3}{x^2}} + \sqrt {1 + \frac{4}{x} + \frac{3}{x^2}}}$As $x 	o \infty$,$\frac{1}{x} 	o 0$ and $\frac{1}{x^2} 	o 0$:$= \frac{4}{\sqrt{1+0+0} + \sqrt{1+0+0}} = \frac{4}{1+1} = \frac{4}{2} = 2$.

$\mathop {\lim }\limits_{x \to \infty } (\sqrt {{x^2} + 8x + 3} - \sqrt {{x^2} + 4x + 3} ) = $

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