Question

$$\lim_{x \to 1} \frac{\sqrt{x + 3} - 2}{x - 1}$$

Explanation:

Step1: Rationalize the numerator

Multiply the numerator and denominator by $\sqrt{x + 3}+2$.

$$\lim_{x ightarrow1}\frac{(\sqrt{x + 3}-2)(\sqrt{x + 3}+2)}{(x - 1)(\sqrt{x + 3}+2)}$$

Using the difference - of - squares formula $(a - b)(a + b)=a^{2}-b^{2}$, the numerator becomes $(x + 3)-4=x - 1$.
So the limit is $\lim_{x
ightarrow1}\frac{x - 1}{(x - 1)(\sqrt{x + 3}+2)}$.

Step2: Simplify the expression

Cancel out the common factor $(x - 1)$ (since $x
eq1$ as we are taking the limit as $x
ightarrow1$).
We get $\lim_{x
ightarrow1}\frac{1}{\sqrt{x + 3}+2}$.

Step3: Substitute $x = 1$

Substitute $x = 1$ into the simplified expression $\frac{1}{\sqrt{x + 3}+2}$.
When $x = 1$, $\sqrt{1+3}+2=\sqrt{4}+2=2 + 2=4$.
So the value is $\frac{1}{4}$.

Answer:

$\frac{1}{4}$