Lim x-> 00 sin(1/x)x^2/(2x-1)?

  • Thread starter cos(e)
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  • #1
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lim x-> 00 sin(1/x)x^2/(2x-1)???

lim x-> 00 sin(1/x)x^2/(2x-1)???

Iv tried sandwich theorem, l'hopitals and i seem to be going nowhere
 

Answers and Replies

  • #2
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Maybe you could use the small angle approximation for sin? (sin(y) = y for small y)
 
  • #3
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But that just makes sin(1/x)=0 which doesnt help, when i graph it i get the limit 1/2 i just dont know how to find it using limit laws
 
  • #4
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You're not applying the small angle approximation. sin(y)=y for small y. I didn't say sin(y)=0 for small y.
 
  • #5
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Oh i see that works, but im not too sure if i am meant to use the small angle approximation, but otherwise thanks
 
  • #6
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Really? I mean it's just an approximation of the taylor series for sin(x). Seems fair game for a calculus class. Sorry I can't be of more help.
 
  • #7
Gib Z
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Personally I don't enjoy justifying that taking sin y = y was accurate enough. It was obvious for cos(e) to question why we were not letting sin y = 0 - You have to take the correct number of terms of the series, and taking more terms until you get the expected answer really isn't good enough imo.

I would have used the Sandwich theorem with [tex] x - \frac{x^3}{6} \leq \sin x \leq x[/tex].
 
  • #8
HallsofIvy
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Since we know that
[tex]\lim_{x\rightarrrow 0}\frac{sin(x)}{x}= 1[/itex]
I would be inclined, as a first step, to let y= 1/x. Then the limit becomes
[tex]\lim_{y\rightarrow 0}\frac{sin(y)}{y^2(2/y- 1)}[/tex]
which, multiplying one of the y's from y2 into the parentheses, gives
[tex]\lim_{y\rightarrow 0}\frac{sin y}{y}\frac{1}{2-y}[/itex]
which should be pretty easy.
 
  • #9
statdad
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Just a question: does "x -> 00" mean the limit as x goes to zero or as x goes to infinity?
 
  • #10
Gib Z
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Since we know that
[tex]\lim_{x\rightarrrow 0}\frac{sin(x)}{x}= 1[/itex]
I would be inclined, as a first step, to let y= 1/x. Then the limit becomes
[tex]\lim_{y\rightarrow 0}\frac{sin(y)}{y^2(2/y- 1)}[/tex]
which, multiplying one of the y's from y2 into the parentheses, gives
[tex]\lim_{y\rightarrow 0}\frac{sin y}{y}\frac{1}{2-y}[/itex]
which should be pretty easy.

A very elegant solution Halls =]

Just a question: does "x -> 00" mean the limit as x goes to zero or as x goes to infinity?

It means the limit as x goes to infinity, commonly used notation on forums for those who don't use latex.
 
  • #11
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derivation to funkton
f(x)= e^sin^22x when x=pi/2
i try but i get "0"... is this right solution?
plz if any one help me
 
  • #12
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derivation to funkton
f(x)= e^sin^22x when x=pi/2
i try but i get "0"... is this right solution?
plz if any one help me
Please start a new thread rather than tacking your question onto a thread that is more than two years old.
 
  • #13
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Please start a new thread rather than tacking your question onto a thread that is more than two years old.

com an man im not so used by sympols in com puter .. but i well try
 
  • #14
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calculate the derivative of the function. specify the exact answer

f(x)= e^sin^2(2x) when x=∏/2
 
  • #15
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I wasn't talking about the symbols - don't use this thread for an unrelated problem. Please start a new thread.
 
  • #16
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Please start a new thread rather than tacking your question onto a thread that is more than two years old.

I wasn't talking about the symbols - don't use this thread for an unrelated problem. Please start a new thread.[/QUO

heheh okok sorry i got u......
 

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