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Find the limit of the this equation

  • Thread starter shwanky
  • Start date
  • #1
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Homework Statement


Evaluate the limit if it exists.


Homework Equations


[tex]lim_{x\to2} \frac{\sqrt{x-6}-2}{\sqrt{x-3}-1}[/tex]


The Attempt at a Solution


Completely lost, I've tried, finding the conjugate of both the numerator or denominator and I'm unable to find a solution.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
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Are you sure that is the problem? AT x= 2, the fraction is
[tex]\frac{2i- 2}{i- 1}= 2[/tex]
Nothing hard about that!

Perhaps you meant
[tex]\frac{\sqrt{6- x}- 2}{\sqrt{3- x}-1}[/itex]
which is "0/0" when x= 2.
Use the "conjugates" as you mention- multiply both numerator and denominator by [itex](\sqrt{x-6}+2)(\sqrt{x-3}+1)[/itex]. You will get x2-4 in the numerator and x-2 in the denominator and can cancel.
 
  • #3
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ok, here's another one I don't understand.

Homework Equations


[tex]lim_{t \to 0} \frac{x+1}{xsin(\Pi x)}[/tex]

The Attempt at a Solution


[tex]lim_{t \to 0} \frac{x+1}{xsin(\Pi x)}(\frac{x-1}{x-1})[/tex]
[tex]lim_{t \to 0} \frac{(x^2-1)}{(x^2-1)sin(\Pi x)}[/tex]
[tex]lim_{t \to 0} \frac{(1)}{sin(\Pi x)}[/tex]

At this point I get stuck... the [tex]sin(\Pi x)[/tex] is still 0. I tried using some of the trig identities, that I actually remembered, and got this.

[tex]lim_{t \to 0} \frac{(sin^2(\Pi x) + cos^2(\Pi x)}{sin(\Pi x)}[/tex]

Now I thought if I could get I could get the numerator in terms of sine, I could simple divide into it, but I'm not sure if this would work :-/... any suggestions?
 
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  • #4
43
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is it possible for me to do

[tex]lim_{t \to 0} \frac{(1)}{sin(\Pi x)}[/tex]
[tex]lim_{t \to 0} \frac{(1)}{csc(\Pi x)}[/tex]
[tex]lim_{t \to 0} sin(\Pi x)[/tex]

?
 
  • #5
43
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mmm never mind. there was a problem in my solution. [tex]xsin(\Pi x)(x-1)[/tex] does not equal [tex](x2-1)sin(\Pi x)[/tex]... I will try again...
 
  • #6
43
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ok, I think I got it.

[tex]lim_{t \to 1} \frac{x+1}{xsin(\Pi x)}[/tex]

[tex]lim_{t \to 1} \frac{x+1}{xsin(\Pi x)}(\frac{csc(\Pi x)}{csc(\Pi x)})[/tex]

[tex]lim_{t \to 1} \frac{(x+1)(csc(\Pi x)}{x}[/tex]

[tex]lim_{t \to 1} \frac{(x+1)(csc(\Pi x)}{x}[/tex]

[tex]lim_{t \to 1} csc(\Pi x) + \frac{(csc(\Pi x)}{x}[/tex]

[tex]lim_{t \to 1} csc(\Pi * 1) + \frac{(csc(\Pi *1)}{1}[/tex]

[tex]lim_{t \to 1} csc(\Pi) + (csc(\Pi)[/tex]

[tex]lim_{t \to 1} 2csc(\Pi)[/tex]

[tex]lim_{t \to 1+} 2csc(\Pi) = +\infty[/tex]

[tex]lim_{t \to 1-} 2csc(\Pi) = -\infty[/tex]
 
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  • #7
43
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Sorry, the limit is t->1+ not t->0...
 
  • #8
43
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ok, now if I take the left hand limit and right hand limit, the right hand limit approaches infinity while the left hand appraches negative infinity?
 
  • #9
HallsofIvy
Science Advisor
Homework Helper
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Does it bother you at all that you are taking the limit as t-> 1 but there is no t in the formula?! (Mathematics requires precision- be careful what you write!)

Perhaps you meant
[tex]lim_{t \to 1^+} \frac{t+1}{tsin(\Pi t)}[/tex]

I don't see difficulty with that. The denominator is continuous and at t= 1 is [itex](1)sin(\pi)= 0[/itex] but the numerator does NOT go to 0. What does that tell you?

(Don't capitalize "Pi"- you want \pi: [itex]\pi[/itex].)
 
  • #10
43
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o_O... I am such a dork...

[tex]\lim_{t \to 1^+} \frac{t+1}{tsin(\pi t)} = -\infty [/tex] as [tex]t \to 1^+[/tex]
 
Last edited:

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