<s>[INST] Is this Function differentiable and continuous at x=0? Is $f(x)$ continuous and differentiable at $x = 0$ ?
$$f(x) = x(\sqrt{x} - \sqrt{x+1})$$
 [/INST] First $f$ is defined only for $x \ge 0$.
Then, you have $$f(x) = x(\sqrt{x} - \sqrt{x+1})=x(\sqrt{x} - \sqrt{x+1})\frac{\sqrt{x} + \sqrt{x+1}}{\sqrt{x} + \sqrt{x+1}} = \frac{-x}{\sqrt{x} + \sqrt{x+1}}$$
which is continuous at $x=0$ with $f(0)=\lim\limits_{x \to 0^+} f(x) = 0$. Moreover $$\frac{f(x)}{x}=\frac{-1}{\sqrt{x} + \sqrt{x+1}}$$ has a right hand side limit as $x$ goes to $0$ equal to $-1$. So $f^\prime(0)=-1$.
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