$$\\lim_{x \\to 1} \\frac{\\sqrt{x + 3} - 2}{x - 1}$$
$$\\lim_{x \\to 1} \\frac{\\sqrt{x + 3} - 2}{x - 1}$$
Answer
Explanation:
Step1: Rationalize the numerator
Multiply the numerator and denominator by $\sqrt{x + 3}+2$. $$\lim_{x\rightarrow1}\frac{(\sqrt{x + 3}-2)(\sqrt{x + 3}+2)}{(x - 1)(\sqrt{x + 3}+2)}$$ 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 $\lim_{x\rightarrow1}\frac{x - 1}{(x - 1)(\sqrt{x + 3}+2)}$.
Step2: Simplify the expression
Cancel out the common factor $(x - 1)$ (since $x\neq1$ as we are taking the limit as $x\rightarrow1$). We get $\lim_{x\rightarrow1}\frac{1}{\sqrt{x + 3}+2}$.
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:
$\frac{1}{4}$