Practicing Success
Practicing Success
CUET Preparation Today
$\underset{x→∞}{\lim}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})$ is given by |
0 $\frac{1}{2}$ $\log 2$ none of these |
$\frac{1}{2}$ |
$\underset{x→∞}{\lim}(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x})=\underset{x→∞}{\lim}\frac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}=\underset{x→∞}{\lim}\frac{\sqrt{1+x^{-1/2}}}{\sqrt{1+\sqrt{x^{-1}+x^{-3/2}}}+1}=\frac{1}{2}$ |
