Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$

Multiply both numerator and denominator by $\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}$

You will get $$\dfrac{\sqrt{x+\sqrt{x}}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}$$

Divide both numerator and denominator by $\sqrt{x}$

$$\dfrac{\sqrt{1+\dfrac{1}{\sqrt{x}}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x\sqrt{x}}}}+1}$$

On finding the limit to infinity, you get

$$\dfrac{\sqrt{1+0}}{\sqrt{1+0}+1} = \dfrac{1}{2}$$

Tags:

Limits

Calculus

Radicals

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