<s>[INST] 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.
 [/INST] Set $\dfrac1x=h^2$ where $h>0$ to get
$$\lim_{ x \to \infty}  \sqrt{x+\sqrt{x+\sqrt{x}}}$$
$$=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}}}}=\sqrt{\frac1{h^2}+\sqrt{\frac1{h^2}+\frac1h}}=\sqrt{\frac1{h^2}+\sqrt{\frac{h+1}{h^2}}}$$
$$=\sqrt{\frac1{h^2}+\frac{\sqrt{h+1}}h}=\sqrt{\frac{1+h\sqrt{h+1}}{h^2}}=\frac{\sqrt{1+h\sqrt{h+1}}}h$$
$$\implies\lim_{ x \to \infty}  \sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}=\lim_{h\to0}\frac{\sqrt{1+h\sqrt{h+1}}-1}h$$
$$=\lim_{h\to0}\frac{1+h\sqrt{h+1}-1}{h(\sqrt{1+h\sqrt{h+1}}+1)}=\frac{\sqrt{0+1}}{\sqrt{1+0\sqrt{0+1}}+1}$$
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