Question

lim(x tends to infinity) root(x+1)-root(x)

Answer 1

Dear Student,
$$\begin{gathered} \underset{x\to \infty }{\lim }(\sqrt{x+1}-\sqrt{x}) \\ Onrationali\sin g,weget, \\ \underset{x\to \infty }{\lim }\frac{(\sqrt{x+1}-\sqrt{x})\times (\sqrt{x+1}+\sqrt{x})}{(\sqrt{x+1}+\sqrt{x})} \\ i.e\underset{x\to \infty }{\lim }\frac{{\left(\sqrt{x+1}\right)}^2-{\left(\sqrt{x}\right)}^2}{(\sqrt{x+1}+\sqrt{x})} \\ i.e\underset{x\to \infty }{\lim }\frac{1}{(\sqrt{x+1}+\sqrt{x})} \\ i.e\underset{x\to \infty }{\lim }\frac{1}{\sqrt{x}(\sqrt{1+{\displaystyle \frac{1}{x}}}+1)} \\ Now,whenx->\infty ,\frac{1}{x}->0(Alsox->\infty ,\sqrt{x}->\infty ) \\ i.e\underset{x\to \infty }{\lim }\frac{1}{\sqrt{x}(\sqrt{1+{\displaystyle \frac{1}{x}}}+1)}=\frac{1}{\infty }\times \frac{1}{\sqrt{1+0}+1}=0\times 1=0 \end{gathered}$$
Regards.