Đáp án: $1$
Giải thích các bước giải:
f.Ta có:
$\lim_{x\to0}\dfrac{\sqrt{1+x}-\sqrt{1+x^2}}{\sqrt{1-x^2}-\sqrt{1-x}}$
$=\lim_{x\to0}\dfrac{\dfrac{(1+x)-(1+x^2)}{\sqrt{1+x}+\sqrt{1+x^2}}}{\dfrac{(1-x^2)-(1-x)}{\sqrt{1-x^2}+\sqrt{1-x}}}$
$=\lim_{x\to0}\dfrac{\dfrac{x-x^2}{\sqrt{1+x}+\sqrt{1+x^2}}}{\dfrac{x-x^2}{\sqrt{1-x^2}+\sqrt{1-x}}}$
$=\lim_{x\to0}\dfrac{\sqrt{1-x^2}+\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1+x^2}}$
$=\dfrac{\sqrt{1-0^2}+\sqrt{1-0}}{\sqrt{1+0}+\sqrt{1+0^2}}$
$=1$
