Q.$\displaystyle\lim_{x\to 0}\left[\frac{\sqrt{1+x^2}-\sqrt{1+x}}{\sqrt{1+x^3}-\sqrt{1+x}}\right]$

Numerator: $\sqrt{1+x^2}-\sqrt{1+x}=\dfrac{(1+x^2)-(1+x)}{\sqrt{1+x^2}+\sqrt{1+x}}=\dfrac{x(x-1)}{\sqrt{1+x^2}+\sqrt{1+x}}$. Denominator: $\sqrt{1+x^3}-\sqrt{1+x}=\dfrac{(1+x^3)-(1+x)}{\sqrt{1+x^3}+\sqrt{1+x}}=\dfrac{x(x-1)(x+1)}{\sqrt{1+x^3}+\sqrt{1+x}}$. Dividing, the shared factor $x(x-1)$ cancels, leaving $\dfrac{\sqrt{1+x^3}+\sqrt{1+x}}{(x+1)\left(\sqrt{1+x^2}+\sqrt{1+x}\right)}$. At $x=0$: $\dfrac{1+1}{1\times(1+1)}=1$.

Rationalize numerator and denominator by their separate conjugates, factor each result, cancel the shared $x(x-1)$, substitute.

Not spotting that $x^3-x=x(x-1)(x+1)$ shares a factor with $x^2-x=x(x-1)$ from the numerator, so failing to cancel correctly.