# How can I calculate left and right-sided limits?

1. Aug 31, 2008

### Phizyk

Hi,
How can I calculate left and right-sided limits?
$$\frac{x}{a}[\frac{b}{x}]$$
$$\frac{b}{x}[\frac{x}{a}]$$
$$\frac{x}{\sqrt{|sinx|}}$$
in point x=0.
Thanks for help.

2. Aug 31, 2008

Re: limits

What have you done? think about how the definition of absolute value and how $$\sin x$$ behaves when $$x \approx 0$$.

3. Aug 31, 2008

### HallsofIvy

Staff Emeritus
Re: limits

For x not equal to 0, this is just b/a and so has b/a as both right and left sided limits.
Or did you mean (x/a)|b/x|? In that case, you take left and right limits by looking at:
If x> 0 then |b/x|= |b|/x so (x/a)(|b|/x)= |b|/a
If x< 0 then |b/x|= -|b|/x so (x/a)(|b|/x)= -|b|/a

The last one should be easy. Since sin(-x)= -sin(x), |sin(-x)|= |sin(x)| and the only difference between x< 0 and x> 0 is in the numerator.

4. Aug 31, 2008

### Phizyk

Re: limits

$$[\frac{b}{x}]$$ it is entier function. I can not solve second case... It is harder than first. Can I do $$(\frac{x}{a}-1)\frac{b}{x}\leq{[\frac{x}{a}]\frac{b}{x}}\leq{\frac{b}{a}}$$ and use $$|f(x)-g|\leq{\epsilon}$$ so $$g=\frac{b}{a}$$?

5. Aug 31, 2008

### HallsofIvy

Staff Emeritus
Re: limits

I don't know what that means.

Where did the "-1" in $$\frac{x}{a}-1[/itex] come from? 6. Aug 31, 2008 ### Phizyk Re: limits [tex][\frac{b}{x}]$$ the floor and ceiling functions.
$$x-1\leq{[x]}\leq{x}$$

7. Sep 1, 2008

### HallsofIvy

Staff Emeritus
Re: limits

Chose one! Does it mean the floor or the ceiling. It can't be both! If you mean [x] is the integer between x-1 and x, then it is the floor.