\(\rm \displaystyle \lim_{x\rightarrow \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\) is equal to:

Concept:

Indeterminate Forms: Any expression whose value cannot be defined, like 0/0, ±∞/∞, 0⁰, ∞⁰ etc.

For the indeterminate form ∞ - ∞, first rationalize by multiplying with the conjugate and then divide the terms by the highest power of the variable to get terms so that 1/x → 0 as x → ∞.

Calculation: 

\(\rm \displaystyle \lim_{x\rightarrow \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)=\infty - \infty\) ∴ we first rationalize it by multiplying it with its conjugate as follows:

\(\rm \displaystyle \lim_{x\rightarrow \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\\=\displaystyle \lim_{x\rightarrow \infty} \dfrac{\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)\left(\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}\right)}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)

\(\rm \displaystyle =\lim_{x\rightarrow \infty} \dfrac{\sqrt{x+\sqrt x}}{\sqrt{x+\sqrt{x+\sqrt{x}}}+\sqrt{x}}\)

\(\rm \displaystyle =\lim_{x\rightarrow \infty} \dfrac{\sqrt{1+\dfrac{1}{\sqrt x}}}{\sqrt{1+\sqrt{\dfrac{1}{x}+\dfrac{1}{x^{3/2}}}}+1}\)                             ...(Dividing by √x)

\(\rm =\dfrac{\sqrt{1+0}}{\sqrt{1+\sqrt{0+0}}+1}=\dfrac{1}{2}\).