How can I calculate left and right-sided limits?

[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.

 

[statdad]

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

 

[HallsofIvy]

Hi,
How can I calculate left and right-sided limits?
\frac{x}{a}[\frac{b}{x}]

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

\frac{b}{x}[\frac{x}{a}]

Same comments

\frac{x}{\sqrt{|sinx|}}
in point x=0.
Thanks for help.

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.

 

[Phizyk]

[\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}?

 

[HallsofIvy]

[\frac{b}{x}] it is entier function.

I don't know what that means.

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}?

Where did the "-1" in \frac{x}{a}-1[/itex] come from?

 

[Phizyk]

[\frac{b}{x}] the floor and ceiling functions.
x-1\leq{[x]}\leq{x}

 

[HallsofIvy]

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.