Proving function Converges

  • #1
PsychonautQQ
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Homework Statement


How do I prove the function ((2^x)-1)sin(y))/(xy) converges to ln(2) in the case of x=0 and y=0..?


Homework Equations





The Attempt at a Solution


Yeah I'm trying to figure out where to start honestly... I know I'm not suppose to post unless I have a real attempt at the solution.. I could BS something but anyone want to point me in the right direction? Squeeze Theorem? Is there some key property I should be aware of?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


How do I prove the function ((2^x)-1)sin(y))/(xy) converges to ln(2) in the case of x=0 and y=0..?


Homework Equations





The Attempt at a Solution


Yeah I'm trying to figure out where to start honestly... I know I'm not suppose to post unless I have a real attempt at the solution.. I could BS something but anyone want to point me in the right direction? Squeeze Theorem? Is there some key property I should be aware of?

Why do you think the answer is ##\ln 2##? Try some special limits to get a feel for it. What if ##y\rightarrow 0## before ##x##, for example.
 
  • #3
PsychonautQQ
784
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the answer is ln(2), the question says that, we just have to PROVE it is.
 
  • #4
dirk_mec1
761
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Yes, the answer is indeed ln(2) I used mclaurin series to prove it.
 
  • #5
pasmith
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Homework Statement


How do I prove the function ((2^x)-1)sin(y))/(xy) converges to ln(2) in the case of x=0 and y=0..?


Homework Equations





The Attempt at a Solution


Yeah I'm trying to figure out where to start honestly... I know I'm not suppose to post unless I have a real attempt at the solution.. I could BS something but anyone want to point me in the right direction? Squeeze Theorem? Is there some key property I should be aware of?

Your function is a function of x only times a function of y only:
[tex]
\frac{(2^x - 1)\sin y}{xy} = \frac{2^x - 1}{x} \frac{\sin y}{y}
[/tex]

One can show that if [itex]\lim_{x \to 0} f(x)[/itex] and [itex]\lim_{y \to 0} g(y)[/itex] exist then
[tex]
\lim_{(x,y) \to 0} f(x)g(y) = \left(\lim_{x \to 0} f(x) \right)\left( \lim_{y \to 0} g(y) \right)
[/tex]

Also remember that
[tex]
\lim_{x \to 0} \frac{f(x) - f(0)}x = f'(0)
[/tex]
assuming the limit exists.
 

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