# Help-limits with integral part

Can you help me to find the limits? $$\lim_{x \to 0^{+}}\frac{x}{a}\left[\frac{b}{x}\right] , \quad \lim_{x \to 0^{+}}\frac{b}{x}\left[\frac{x}{a}\right], \quad a,b>0$$ And what happens when $x \to 0^{-}$ ?

• What does $[\frac{b}{x}]$ mean, is that a special notation for something? Feb 10, 2014 at 22:48
• I guess it's the biggest integer
– mjb4
Feb 10, 2014 at 22:49
• It is the integral part!!! Feb 10, 2014 at 22:50

For the first: we have $$\Bigl[\frac{b}{x}\Bigr]=\frac{b}{x}-\varepsilon(x)\ ,$$ where $0\le\varepsilon(x)<1$. Hence $$\frac{x}{a}\Bigl[\frac{b}{x}\Bigr]=\frac{b}{a}-\frac{x}{a}\varepsilon(x) \to\frac{b}{a}$$ as $x\to0^+$, beacuse the second term lies between $0$ and $x/a$.

For the second, as soon as $x<a$ we have $[x/a]=0$ and so $$\frac{b}{x}\Bigl[\frac{x}{a}\Bigr]=0\to0$$ as $x\to0^+$.

• Since we know that $[x]\leq x \geq [x]+1$,could we use this relation instead of the one that you used?? I applied it and got: $$\frac{b}{a}-\frac{x}{a} \leq \frac{x}{a}[\frac{b}{x}] \geq \frac{b}{a}$$..Can we say that the first limit is $0^{+}$ ..And,what's about $x\to O^{-}$.Is this the same result?? Feb 10, 2014 at 23:07
• The second $\ge$ in your comment should be $<$. If you fix this it should be OK. I haven't looked at the $x\to0^-$ case. The same ideas should work, though the answers could possibly be different because for negative $x$ the integer part rounds away from $0$. For example $[\frac{3}{2}]$ is $1$ but $[-\frac{3}{2}]$ is $-2$, not $-1$. See what you can do. Feb 10, 2014 at 23:21
• Using the same relation for the second limit,I have found the following: $$\frac{b}{a}-\frac{b}{x}\leq \frac{b}{x}[\frac{x}{a}] \leq \frac{b}{a}$$ What does this mean??That the limit does not exist???Or have I done something wrong?? Feb 10, 2014 at 23:35
• If you are trying to use the Sandwich Theorem (Pinching Theorem) it will not apply here and you will have to use other methods. Feb 10, 2014 at 23:57
• I got stuck right now...Could you give me a hint?? Feb 11, 2014 at 0:01

This Limit does not exists!

Just look at the subseries $x_n=n$ and $x'_n := 0.9+n$

• Which limit do you mean that does not exist???the second one?? Feb 10, 2014 at 23:10
• hmm I guess both! let me see
– mjb4
Feb 10, 2014 at 23:11
• Could you check what I have done? Since we know that $[x]\leq x \leq [x]+1$,could we use this relation instead of the one that you used?? I applied it and got: $$\frac{b}{a}-\frac{x}{a} \leq \frac{x}{a}[\frac{b}{x}] \geq \frac{b}{a}$$..Can we say that the first limit is $0^{+}$ ..And,what's about $x\to O^{-}$.Is this the same result?? Feb 10, 2014 at 23:13
• ah sorry im totally wrong! I did not see $x\to 0$, they both exists. Dont make it so complex! wait i edit my post
– mjb4
Feb 10, 2014 at 23:16