Question

find the following limit or state that it does not exist.
\\( \lim \limits_{x \to 2} \frac{x - 2}{\sqrt{4x + 1} - 3} \\)

simplify the given limit.
\\( \lim \limits_{x \to 2} \frac{x - 2}{\sqrt{4x + 1} - 3} = \lim \limits_{x \to 2} (\square) \\) (simplify your answer.)

Explanation:

Step1: Rationalize the denominator

Multiply numerator and denominator by the conjugate of the denominator, which is $\sqrt{4x + 1} + 3$.

$$\lim_{x ightarrow 2}\frac{x - 2}{\sqrt{4x + 1}-3}\cdot\frac{\sqrt{4x + 1}+3}{\sqrt{4x + 1}+3}$$

Step2: Simplify the denominator

Using the difference of squares formula $(a - b)(a + b)=a^2 - b^2$, the denominator becomes $(\sqrt{4x + 1})^2-3^2 = 4x + 1 - 9=4x - 8 = 4(x - 2)$.
The numerator is $(x - 2)(\sqrt{4x + 1}+3)$.
So the expression simplifies to:

$$\lim_{x ightarrow 2}\frac{(x - 2)(\sqrt{4x + 1}+3)}{4(x - 2)}$$

Step3: Cancel out common factors

Cancel out the common factor $(x - 2)$ (since $x
ightarrow 2$ but $x
eq2$ at the limit point, we can do this).

$$\lim_{x ightarrow 2}\frac{\sqrt{4x + 1}+3}{4}$$

Answer:

$\frac{\sqrt{4x + 1}+3}{4}$