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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

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} ) = $

Similar Questions

For each $t \in \mathbb{R}$,let $[t]$ be the greatest integer less than or equal to $t$. Then $\lim_{x \to 1^+} \frac{(1 - |x| + \sin |1 - x|) \sin (\frac{\pi}{2} [1 - x])}{|1 - x|^2}$ is:

$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 2}}{{x + 1}}} \right)^{x + 3}}$ is

$\lim _{x \rightarrow 0} \frac{\sqrt{1+x^2}-\sqrt{1-x+x^2}}{3^x-1}$ is equal to

$\mathop {\lim }\limits_{x \to \infty } \sqrt {\frac{{x + \sin x}}{{x - \cos x}}} = $

$\lim _{n \rightarrow \infty} \frac{3 \cdot 2^{n+1}-4 \cdot 5^{n+1}}{5 \cdot 2^{n}+7 \cdot 5^{n}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module