Limits with sine and square roots

  • Thread starter mtayab1994
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  • #1
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Homework Statement



solve the limit: [tex]\lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}[/tex]




The Attempt at a Solution



Is there a way of how i can solve this without using l'hospital's rule or taylor series?
 

Answers and Replies

  • #2
Ray Vickson
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Homework Statement



solve the limit: [tex]\lim_{x\rightarrow1}\frac{sin(x-1)}{\sqrt{x}-1}[/tex]




The Attempt at a Solution



Is there a way of how i can solve this without using l'hospital's rule or taylor series?
What is wrong with using l'Hospital's rule or Taylor series? Of course, changing to the variable [itex] t = \sqrt{x}[/itex] simplifies the question a lot.

RGV
 
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  • #3
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Well i haven't learned l'hospitals rule or taylor's series so i can't use either of them.
 
  • #4
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Multiply numerator and denominator by [itex]\sqrt{x}+1[/itex].
 
  • #5
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Multiply numerator and denominator by [itex]\sqrt{x}+1[/itex].
ok i multiplied by √x+1 and i got:

[tex]\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}[/tex]

now should i cancel the x-1?
 
  • #6
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Do you know the limit

[tex]\lim_{x\rightarrow 0}\frac{\sin(x)}{x}[/tex]
 
  • #7
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Do you know the limit

[tex]\lim_{x\rightarrow 0}\frac{\sin(x)}{x}[/tex]
yes it's 1
 
  • #8
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Now, use that to calculate the limit.
 
  • #9
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alright got it now:

[tex]\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2[/tex]

Is that correct?
 
  • #10
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alright got it now:

[tex]\lim_{x\rightarrow1}\frac{sin(x)}{x}*(\sqrt{x}+1)=2[/tex]

Is that correct?
No.
[tex]\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1[/tex]

Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
[tex]\lim_{y \to 0}\frac{sin(y)}{y} = 1[/tex]
 
  • #11
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No.
[tex]\lim_{x\to 1}\frac{sin(x)}{x}=\frac{sin(1)}{1} \neq 1[/tex]

Split your limit into two limits, and then do some fiddling with the first one so that you can use this limit:
[tex]\lim_{y \to 0}\frac{sin(y)}{y} = 1[/tex]
We still haven't learned how to split a limit into 2 different limits yet.
 
  • #12
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Are you sure? It's a very basic property of limits, and one that is presented pretty early.
[tex]\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)[/tex]

The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.
 
  • #13
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Are you sure? It's a very basic property of limits, and one that is presented pretty early.
[tex]\lim_{x \to a} f(x)\cdot g(x) = \lim_{x \to a}f(x) \cdot \lim_{x \to a}g(x)[/tex]

The above is true as long as both limits on the right exist. As I see it, you pretty much need to use this idea in your problem.
Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.
 
  • #14
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Yes I know that, but we haven't reached it yet in the lesson hence i can't use it.
Then I'm afraid you can't solve this question.

You can't even solve questions like [itex]\lim_{x\rightarrow 1}{x^2}[/itex] in this case.
 
  • #15
SammyS
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ok i multiplied by √x+1 and i got:

[tex]\lim_{x\rightarrow1}\frac{sin(x-1)(\sqrt{x}+1)}{x-1}[/tex]

now should i cancel the x-1?
Do you know the limit

[tex]\lim_{x\rightarrow 0}\frac{\sin(x)}{x}[/tex]
yes it's 1
So, if [itex]\displaystyle \lim_{x\rightarrow 0}\frac{\sin(x)}{x}=1\,,[/itex] then what is [itex]\displaystyle \lim_{x-1\rightarrow 0}\frac{\sin(x-1)}{x-1}\,?[/itex]
 

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