The value of mathop {lim }limits_{x to infty | vedclass

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

Question

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$,we rationalize the expressio

Vedclass · Accepted Answer

$\frac{1}{2}$

Vedclass · Answer

(A) To evaluate the limit $\mathop {\lim }\limits_{x 	o \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$,we rationalize the expression:$= \mathop {\lim }\limits_{x 	o \infty } \frac{(\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})(\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1})}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{({a^2}{x^2} + ax + 1) - ({a^2}{x^2} + 1)}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$$= \mathop {\lim }\limits_{x 	o \infty } \frac{ax}{\sqrt {{a^2}{x^2} + ax + 1} + \sqrt {{a^2}{x^2} + 1}}$Dividing the numerator and denominator by $x$ (assuming $x > 0$ as $x 	o \infty$):$= \mathop {\lim }\limits_{x 	o \infty } \frac{a}{\sqrt {{a^2} + \frac{a}{x} + \frac{1}{{{x^2}}}} + \sqrt {{a^2} + \frac{1}{{{x^2}}}}}$As $x 	o \infty$,$\frac{1}{x} 	o 0$ and $\frac{1}{{{x^2}}} 	o 0$:$= \frac{a}{\sqrt {{a^2}} + \sqrt {{a^2}}} = \frac{a}{2|a|}$If $a > 0$,the limit is $\frac{a}{2a} = \frac{1}{2}$. If $a  0$ based on the options,the answer is $\frac{1}{2}$.

The value of $\mathop {\lim }\limits_{x \to \infty } (\sqrt {{a^2}{x^2} + ax + 1} - \sqrt {{a^2}{x^2} + 1})$ is

Similar Questions

If $S_1 = \sum_{r=1}^{n} r$,$S_2 = \sum_{r=1}^{n} r^2$,and $S_3 = \sum_{r=1}^{n} r^3$,then the value of $\lim_{n \rightarrow \infty} \frac{S_1(1 + \frac{S_3}{4})}{S_2^2}$ is

The value of $\mathop {\lim }\limits_{x \to 2} \frac{{\sqrt {1 + \sqrt {2 + x} } - \sqrt 3 }}{{x - 2}}$ is

$\mathop {\lim }\limits_{n \to \infty } \,\frac{{\sum\limits_{r = 0}^n {{{\tan }^{ - 1}}\left( {1 + r + {r^2}} \right)} }}{n}$ is equal to

$\mathop {\lim }\limits_{n \to \infty } \frac{{n{{(2n + 1)}^2}}}{{(n + 2)({n^2} + 3n - 1)}} = $

$\mathop {\lim }\limits_{x \to 0} \left( \frac{x}{\tan^{-1} 2x} \right) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module