lim _{x rightarrow infty}left[sqrt{x^2+a x+b} | vedclass

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

Question

(C) To evaluate the limit $\lim _{x \rightarrow \infty}\left(\sqrt{x^2+a x+b}-x\right)$,we rationalize the expression:$= \lim _{x}$ ${\rightarrow \inf

Vedclass · Accepted Answer

depends only on $a$

Vedclass · Answer

(C) To evaluate the limit $\lim _{x ightarrow \infty}\left(\sqrt{x^2+a x+b}-xight)$,we rationalize the expression:$= \lim _{x}$ ${ightarrow \infty}\left(\left(\sqrt{x^2+a x+b}-xight) 	imes \frac{\sqrt{x^2+a x+b}+x}{\sqrt{x^2+a x+b}+x}ight)$$= \lim _{x ightarrow \infty}\left(\frac{x^2+a x+b-x^2}{\sqrt{x^2+a x+b}+x}ight)$$= \lim _{x ightarrow \infty}\left(\frac{a x+b}{\sqrt{x^2+a x+b}+x}ight)$Divide the numerator and denominator by $x$:$= \lim _{x ightarrow \infty} \frac{a+\frac{b}{x}}{\sqrt{1+\frac{a}{x}+\frac{b}{x^2}}+1}$As $x ightarrow \infty$,$\frac{1}{x} ightarrow 0$ and $\frac{1}{x^2} ightarrow 0$,so the limit becomes:$= \frac{a+0}{\sqrt{1+0+0}+1} = \frac{a}{2}$Since the result is $\frac{a}{2}$,the limit depends only on $a$.

$\lim _{x \rightarrow \infty}\left[\sqrt{x^2+a x+b}-x\right]$ where $a < 0 < b$.

Similar Questions

$\mathop {\lim }\limits_{x \to 0} {\left( {\frac{{1 + 5{x^2}}}{{1 + 3{x^2}}}} \right)^{1/{x^2}}} = $

Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots \left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}$ for $n \geq 2$. Then, the value of $\lim _{n \rightarrow \infty} x_{n}$ is

If $f(x)$ satisfies $97 f(x) + m f\left(\frac{1}{x}\right) = 0$,where $f(x) = \lim_{n \rightarrow \infty} n(x^{1/n} - 1)$ for $x > 0$,then the value of $m$ is:

If $a > 0, b > 0$,then $\lim _{n \rightarrow \infty}\left(\frac{a + b^{1 / n} - 1}{a}\right)^n =$

$\mathop {\lim }\limits_{x \to 1} \frac{{x + {x^2} + ...... + {x^n} - n}}{{x - 1}}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module