Show $\lim_\limits{n\rightarrow \infty} \sqrt{n}(|c+\frac{d}{\sqrt{n}}|-|c|) = d* sign(c)$

Question

Consider the sequence of real numbers $\sqrt{n}\left(\left|c+\frac{d}{\sqrt{n}}\right|-|c|\right)$ with $c,d \in \mathbb{R}$ and $c\neq 0$. Could you help me to show that

$$\lim_\limits{n\rightarrow \infty} \sqrt{n}\left( \left| c+\frac{d}{\sqrt{n}}\right|-|c|\right) = d\cdot\text{sign}(c)$$

where $\text{sign}(\cdot)$ is the sign (or signum) function defined here. I don't understand what about the case in which $c+\frac{d}{\sqrt{n}}\geq 0$ and $c<0$. Do I need other conditions to establish the result?

Answer 1

A different approach than in user296113's solution:

$$ \sqrt{n}\left(\left\lvert c+\frac{d}{\sqrt{n}}\right\rvert - \lvert c\rvert\right) =\lvert c\rvert \sqrt{n}\left(\left\lvert 1+\frac{d}{c\sqrt{n}}\right\rvert - 1\right) $$ since $\frac{d}{c\sqrt{n}}\xrightarrow[n\to\infty]{} 0$, we have $1+\frac{d}{c\sqrt{n}} > 0$ for $n$ large enough, and then we get $$ \sqrt{n}\left(\left\lvert c+\frac{d}{\sqrt{n}}\right\rvert - \lvert c\rvert\right) =\lvert c\rvert \sqrt{n}\left( 1+\frac{d}{c\sqrt{n}} - 1\right) =\lvert c\rvert \sqrt{n}\frac{d}{c\sqrt{n}} = \frac{\lvert c\rvert}{c}d = \operatorname{sign}(c)d. $$

Note that this means the quantity not only converges to $\operatorname{sign}(c)d$, but is actually equal to it for $n$ big enough. (There exists $N=N(c,d)\geq 1$ such that, for all $n\geq N$, we have equality.)

Answer 2

Multiply by the conjugate you get

$$\frac{2cd+\frac{d^2}{\sqrt n}}{\left|c+\frac d{\sqrt n}\right|+|c|}\xrightarrow{n\to\infty}\frac{c}{|c|}d$$

Answer 3

For $n\gg 0$, $|\frac d{\sqrt n}|<|c|$ and hence $\operatorname{sgn}(c+\frac d{\sqrt n})=\operatorname{sgn}(c)$. So for such $n$ $$ \begin{align}\sqrt n\left(\bigl|c+\tfrac d{\sqrt n}\bigr|-|c|\right)&=\sqrt n\left(\bigl(c+\tfrac d{\sqrt n}\bigr)\cdot \operatorname{sgn}\bigl(c+\tfrac d{\sqrt n}\bigr)-c\cdot\operatorname{sgn}(c)\right) \\&=\sqrt n\left(\bigl(c+\tfrac d{\sqrt n}\bigr)\cdot \operatorname{sgn}(c)-c\cdot\operatorname{sgn}(c)\right)\\ &=\sqrt n\cdot\bigl(c+\tfrac d{\sqrt n}-c\bigr)\cdot\operatorname{sgn}(c)\\ &=d\cdot\operatorname{sgn}(c)\end{align}$$