220. $\\lim_{x\\to 1}\\frac{x^{2}-1}{\\sqrt{x}-1}$

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt{x}+1$. $$\lim_{x\rightarrow1}\frac{(x^{2}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}$$ Since $(\sqrt{x}-1)(\sqrt{x}+1)=x - 1$, the limit becomes $\lim_{x\rightarrow1}\frac{(x^{2}-1)(\sqrt{x}+1)}{x - 1}$.

Step2: Factor the numerator

Factor $x^{2}-1$ as $(x - 1)(x + 1)$. $$\lim_{x\rightarrow1}\frac{(x - 1)(x + 1)(\sqrt{x}+1)}{x - 1}$$ Cancel out the $(x - 1)$ terms (for $x\neq1$, which is valid as we are taking the limit as $x\rightarrow1$, not evaluating at $x = 1$). We get $\lim_{x\rightarrow1}(x + 1)(\sqrt{x}+1)$.

Step3: Substitute $x = 1$

Substitute $x = 1$ into $(x + 1)(\sqrt{x}+1)$. $(1 + 1)(\sqrt{1}+1)=(2)(2)$.

Answer:

$4$