Question

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)

Answer 1

\(\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}+\sqrt{x+4}-3}{x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{\sqrt{x+1}-1+\sqrt{x+4}-2}{x}\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{x+1-1}{\sqrt{x+1}+1}+\dfrac{x+4-4}{\sqrt{x+4}+2}\right)\cdot\dfrac{1}{x}\)

\(=\lim\limits_{x\rightarrow0}\left[\left(\dfrac{x}{\sqrt{x+1}+1}+\dfrac{x}{\sqrt{x+4}+2}\right)\cdot\dfrac{1}{x}\right]\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{\sqrt{x+1}+1}+\dfrac{1}{\sqrt{x+4}+2}\right)\)

\(=\dfrac{1}{\sqrt{0+1}+1}+\dfrac{1}{\sqrt{0+4}+2}=\dfrac{1}{2}+\dfrac{1}{4}=\dfrac{3}{4}\)