# limit of an absolute value function

how would I go about finding the limit of the following absolute value function as it goes to infinity

$\displaystyle\left|\frac{1}{x} - \frac{1}{y}\right|$

Ive never dealt with multivariable limit functions before?

• "...as it goes to infinity[?]" - as what goes to infinity? – The Chaz 2.0 Oct 21 '13 at 22:28
• i picked this problem out of a section on metrics, i have found that it is indeed a metric but must now show where it converges to. I was assuming as it went to infinity since I used the isometry of [0.1) but the original problem states that it is in the half openinterval of (0,1] – cele Oct 21 '13 at 22:32
• I'm having a hard time understanding what you're saying. Are you trying to find $\lim_{x,y \to \infty} |1/x - 1/y|$? – Tyler Oct 21 '13 at 22:39
• the function looks like d(x,y)=abs{(1/x)-(1/y)} – cele Oct 21 '13 at 22:44
• I know that (1/x) converges to 0 so (1/x) and (1/y) would both go to 0. but I was told that my assumption that the limit is 0 was incorrect – cele Oct 21 '13 at 22:46

Claim: $$\lim_{(x,y)\to\infty}\left|\frac{1}{x}-\frac{1}{y}\right|=0$$
Justification: I take $$\lim_{ x \to\infty} \frac{1}{|x|} =0 \tag{1}$$ as known. The double limit $$\lim_{ (x,y) \to\infty} \frac{1}{|x|} =0 \tag{2}$$ follows from (1): if both $x,y$ are large, then $x$ is large. Similarly to (2), $$\lim_{ (x,y) \to\infty} \frac{1}{|y|} =0 \tag{3}$$ Add (2) and (3): $$\lim_{(x,y)\to\infty}\left|\frac{1}{x}\right|+\left|\frac{1}{y}\right|=0$$ The squeeze (based on triangle inequality) finishes the proof: $$0\le \left|\frac{1}{x}-\frac{1}{y}\right|\le \left|\frac{1}{x}\right|+\left|\frac{1}{y}\right|$$