Question

${lim}_{x\to 0}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$

Answer 1

${lim}_{x\to 0}\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}$ [It is of $\left(\frac{0}{0}\right)$ form]

Rationalising the numerator and denominator

$={lim}_{x\to 0}\frac{(\sqrt{1+x^2}-\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})}\times \frac{(\sqrt{1+x^2}+\sqrt{1+x})}{(\sqrt{1+x^3}+\sqrt{1+x})}$

$={lim}_{x\to 0}\frac{\left[\right(1+x^2)-(1+x\left)\right](\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^3}-\sqrt{1+x})(\sqrt{1+x^2}+\sqrt{1+x})(\sqrt{1+x^3}+\sqrt{1+x})}={lim}_{x\to 0}\frac{(x^2-x)(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^2}+\sqrt{1+x})(1+x^3-1-x)}$

$={lim}_{x\to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^2}+\sqrt{1+x})(x^2-1)x}$ [It is of $\left(\frac{0}{0}\right)$ form]

$={lim}_{x\to 0}\frac{x(x-1)(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^2}+\sqrt{1+x})\left(x\right)(x-1)(x+1)}$

$={lim}_{x\to 0}\frac{(\sqrt{1+x^3}+\sqrt{1+x})}{(\sqrt{1+x^2}+\sqrt{1+x})(x+1)}$

$=\frac{2}{2}=1$