1
$\begingroup$

I'm trying to find the limit of $f_{n}(x) =\frac{x}{\sqrt{x^2 +\frac{1}{n}}}$ as $n\rightarrow \infty$ for a fixed $x\in \mathbb{R}$.

If $x=0$, then the limit equals $0$. Otherwise, it should converge to $1$ or $-1$. To prove this formally, though, is giving me trouble.

That is, for $x>0$ and $\epsilon >0$, I want some $N\in \mathbb{N}$ such that for any $n\geq N$, $\big|\frac{x}{\sqrt{x^2 +\frac{1}{n}}} -1\big|<\epsilon$. I'm really not sure how to find this $N$. I know that it can depend on $\epsilon$ or $x$. What's a good way to approach the algebra here? I've tried factoring the $x^2$ out, but I didn't get anywhere.

  • 1
    If $x>0$ I think this converges to $1$ and if $x<0$ then I think this converges to $-1$. – 2017-01-11
  • 0
    Split it into cases for $x<0$, $x=0$, and $x>0$. – 2017-01-11

1 Answers 1

1

For $x>0$ we have:

$$\left\lvert\frac{x}{\sqrt{x^{2}+\frac{1}{n}}}-1\right\rvert=\left\lvert\frac{x-\sqrt{x^{2}+\frac{1}{n}}}{\sqrt{x^{2}+\frac{1}{n}}}\right\rvert=\left\lvert\frac{\frac{1}{n}}{\sqrt{x^{2}+\frac{1}{n}}\left(x+\sqrt{x^{2}+\frac{1}{n}}\right)}\right\rvert\le\frac{\frac{1}{n}}{2x^{2}}$$

The manipulations should be similar for $x<0$.