Find the value of : $\lim_{ x \to \infty} \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$ I need to calculate the limit of the function below:
$$\lim_{ x \to \infty}  \left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right)$$
I tried multiplying by the conjugate, substituting $x=\frac{1}{t^4}$, and both led to nothing.

A useful result for a brute force solution to these sorts of computations is $\sqrt{1+ \theta} = 1+ {1 \over 2} \theta + r (\theta)$, where $\lim_{\theta \to 0} { r(\theta)\over \theta} = 0$
Note that $\sqrt{x+\sqrt{x+ \sqrt{x}}} = \sqrt{x} \sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}}$, and so
$\sqrt{x+\sqrt{x+ \sqrt{x}}} - \sqrt{x} = \sqrt{x} (\sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}} -1)  $.
Hence
$d(x)=\sqrt{x} (\sqrt{1+ \sqrt{{1 \over x}+\sqrt{1 \over x^3}
}} -1) = \sqrt{x}({1 \over 2}\sqrt{{1 \over x}+\sqrt{1 \over x^3}}+r(\sqrt{{1 \over x}+\sqrt{1 \over x^3}}))$.
Simplifying gives
$d(x) = {1 \over 2} \sqrt{1+\sqrt{1 \over x}}+{1 \over 2} {r(\sqrt{{1 \over x}+\sqrt{1 \over x^3}}) \over \sqrt{1 \over x}}$.
Hence $\lim_{x \to \infty} d(x) = {1 \over 2}$.
