Question

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

Explanation:

Step1: Rationalize the denominator

Multiply the numerator and denominator by $\sqrt{x}+1$.

$$\lim_{x ightarrow1}\frac{(x^{2}-1)(\sqrt{x}+1)}{(\sqrt{x}-1)(\sqrt{x}+1)}$$

Since $(\sqrt{x}-1)(\sqrt{x}+1)=x - 1$, the limit becomes $\lim_{x
ightarrow1}\frac{(x^{2}-1)(\sqrt{x}+1)}{x - 1}$.

Step2: Factor the numerator

Factor $x^{2}-1$ as $(x - 1)(x + 1)$.

$$\lim_{x ightarrow1}\frac{(x - 1)(x + 1)(\sqrt{x}+1)}{x - 1}$$

Cancel out the $(x - 1)$ terms (for $x
eq1$, which is valid as we are taking the limit as $x
ightarrow1$, not evaluating at $x = 1$).
We get $\lim_{x
ightarrow1}(x + 1)(\sqrt{x}+1)$.

Step3: Substitute $x = 1$

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

Answer:

$4$