Question

$L{t}_{x\to 0}=\frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1-x^2}-\sqrt{1-x}}=$

• A. $1$
• B. $-1$
• C. $0$
• D. $2$

Answer 1

Answer: (A)

$1$

$=L{t}_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1-x^2}-\sqrt{1-x}}$
$=L{t}_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1-x^2}-\sqrt{1-x}}\frac{\sqrt{1+x}+\sqrt{1+x^2}}{\sqrt{1+x}+\sqrt{1+x^2}}\frac{\sqrt{1-x^2}+\sqrt{1-x}}{\sqrt{1-x^2}+\sqrt{1-x}}$
$=L{t}_{x\to 0}\frac{x-x^2}{x-x^2}\frac{\sqrt{1-x^2}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1+x^2}}$
$=\frac{2}{2}=1$
Alternative Method:
$L{t}_{x\to 0}\frac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1-x^2}-\sqrt{1-x}}$
It is of the form $\frac{0}{0}$, so applying L-Hospital's rule,
$=\underset{x\to 0}{lim}\frac{\frac{1}{2\sqrt{1+x}}-\frac{2x}{2\sqrt{1+x^2}}}{\frac{-2x}{2\sqrt{1-x^2}}-\frac{-1}{2\sqrt{1-x}}}$
$=1$