How can I calculate left and right-sided limits?

  • Thread starter Phizyk
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  • #1
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Hi,
How can I calculate left and right-sided limits?
[tex]\frac{x}{a}[\frac{b}{x}][/tex]
[tex]\frac{b}{x}[\frac{x}{a}][/tex]
[tex]\frac{x}{\sqrt{|sinx|}}[/tex]
in point x=0.
Thanks for help.
 

Answers and Replies

  • #2
statdad
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What have you done? think about how the definition of absolute value and how [tex] \sin x [/tex] behaves when [tex] x \approx 0 [/tex].
 
  • #3
HallsofIvy
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Hi,
How can I calculate left and right-sided limits?
[tex]\frac{x}{a}[\frac{b}{x}][/tex]
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

[tex]\frac{b}{x}[\frac{x}{a}][/tex]
Same comments

[tex]\frac{x}{\sqrt{|sinx|}}[/tex]
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.
 
  • #4
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[tex][\frac{b}{x}][/tex] it is entier function. I can not solve second case... It is harder than first. Can I do [tex](\frac{x}{a}-1)\frac{b}{x}\leq{[\frac{x}{a}]\frac{b}{x}}\leq{\frac{b}{a}}[/tex] and use [tex]|f(x)-g|\leq{\epsilon}[/tex] so [tex]g=\frac{b}{a}[/tex]?
 
  • #5
HallsofIvy
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[tex][\frac{b}{x}][/tex] 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 [tex](\frac{x}{a}-1)\frac{b}{x}\leq{[\frac{x}{a}]\frac{b}{x}}\leq{\frac{b}{a}}[/tex] and use [tex]|f(x)-g|\leq{\epsilon}[/tex] so [tex]g=\frac{b}{a}[/tex]?
Where did the "-1" in [tex]\frac{x}{a}-1[/itex] come from?
 
  • #6
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[tex][\frac{b}{x}][/tex] the floor and ceiling functions.
[tex]x-1\leq{[x]}\leq{x}[/tex]
 
  • #7
HallsofIvy
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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.
 

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