Q84 / 180 · COMEDK 2015 · Mathematics

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

  • Correct answer
    A
    $\frac{b+1}{2 a}$
  • B
    $\frac{b}{a}$
  • C
    0
  • D
    $\frac{2 b}{a}$

Solution

We have, $\lim _{x \rightarrow \infty}\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right)$ $$ \begin{aligned} &=\lim _{x \rightarrow \infty}\left(\left(\sqrt{a^{2} x^{2}+b x+x}-a x\right) \times \frac{\left(\sqrt{a^{2} x^{2}+b x+x}+a x\right)}{\left(\sqrt{a^{2} x^{2}+b x}+x+a x\right)}\right) \\ &=\lim _{x \rightarrow \infty}\left(\frac{\left(a^{2} x^{2}+b x+x\right)-a^{2} x^{2}}{x\left\{\sqrt{a^{2}+\frac{b}{x}+\frac{1}{x}}+a\right\}}\right) \\ &=\lim _{x \rightarrow \infty}\left(\frac{(b+1) x}{x\left\{\sqrt{a^{2}+\frac{b}{x}+\frac{1}{x}}+a\right\}}\right)=\frac{b+1}{2 a} \end{aligned} $$

Hence, the correct answer is option: 'A'.

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