MCQ
$\lim _{x \rightarrow 1} \frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}$ is equal to
  • A
    $\frac{1}{2}$
  • B
    $\sqrt{2}+1$
  • C
    1
  • $1+\frac{1}{\sqrt{2}}$

Answer

Correct option: D.
$1+\frac{1}{\sqrt{2}}$
(D)
$\lim _{x \rightarrow 1} \frac{\sqrt{x^2-1}+\sqrt{x-1}}{\sqrt{x^2-1}}$
$=\lim _{x \rightarrow 1} \frac{\sqrt{(x-1)(x+1)}+\sqrt{x-1}}{\sqrt{(x-1)(x+1)}}$
$=\lim _{x \rightarrow 1} \frac{[\sqrt{x+1}+1]}{\sqrt{x+1}}$
$=\frac{\sqrt{2}+1}{\sqrt{2}}$
$=1+\frac{1}{\sqrt{2}}$

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