Question

\(\lim\limits_{x\rightarrow2}\dfrac{\sqrt{4x+1}-3}{x^2-4}\)

Answer 1

\(\lim\limits_{x\rightarrow2}\)\(\dfrac{\sqrt{4x+1-3}}{x^2-4}\) = \(\lim\limits_{x\rightarrow2}\)\(\dfrac{\left(\sqrt{4x+1-3}\right)\left(\sqrt{4x+1+3}\right)}{\left(x^2-4\right)\left(\sqrt{4x+1+3}\right)}\)

=\(\lim\limits_{x\rightarrow2}\)\(\dfrac{4x+1-9}{\left(x^2-4\right)\left(\sqrt{4x+1+3}\right)}\) = \(\lim\limits_{x\rightarrow2}\)\(\dfrac{4x-8}{\left(x^2-4\right)\left(\sqrt{4x+1+3}\right)}\)

=\(\lim\limits_{x\rightarrow2}\)\(\dfrac{4}{\left(x+2\right)\left(\sqrt{4x+1+3}\right)}\)=\(\dfrac{1}{6}\)

 

Answer 2