Solve the Following Questions


If $\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}-a x\right)=b$, then the ordered pair $(\mathrm{a}, \mathrm{b})$ is:

  1. $\left(1, \frac{1}{2}\right)$

  2. $\left(1,-\frac{1}{2}\right)$

  3. $\left(-1, \frac{1}{2}\right)$

  4. $\left(-1,-\frac{1}{2}\right)$

Correct Option: , 2


$\lim _{x \rightarrow \infty}\left(\sqrt{x^{2}-x+1}\right)-a x=b \quad(\infty-\infty)$

$\Rightarrow a>0$

Now, $\lim _{x \rightarrow \infty} \frac{\left(x^{2}-x+1-a^{2} x^{2}\right)}{\sqrt{x^{2}-x+1}+a x}=b$

$\Rightarrow \lim _{x \rightarrow \infty} \frac{\left(1-a^{2}\right) x^{2}-x+1}{\sqrt{x^{2}-x+1}+a x}=b$

$\Rightarrow \lim _{x \rightarrow \infty} \frac{\left(1-a^{2}\right) x^{2}-x+1}{x\left(\sqrt{1-\frac{1}{x}+\frac{1}{x^{2}}}+a\right)}=b$

$\Rightarrow 1-\mathrm{a}^{2}=0 \Rightarrow \mathrm{a}=1$

Now, $\lim _{x \rightarrow \infty} \frac{-x+1}{x\left(\sqrt{\left.1-\frac{1}{x}+\frac{1}{x^{2}}+a\right)}\right.}=b$

$\Rightarrow \frac{-1}{1+a}=b \Rightarrow b=-\frac{1}{2}$

$(a, b)=\left(1,-\frac{1}{2}\right)$

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