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

[cos(e)]

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'hospital's and i seem to be going nowhere

 

[jhicks]

Maybe you could use the small angle approximation for sin? (sin(y) = y for small y)

 

[cos(e)]

But that just makes sin(1/x)=0 which doesn't help, when i graph it i get the limit 1/2 i just don't know how to find it using limit laws

 

[jhicks]

You're not applying the small angle approximation. sin(y)=y for small y. I didn't say sin(y)=0 for small y.

 

[cos(e)]

Oh i see that works, but I am not too sure if i am meant to use the small angle approximation, but otherwise thanks

 

[jhicks]

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.

 

[Gib Z]

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 $$x - \frac{x^3}{6} \leq \sin x \leq x$$.

 

[HallsofIvy]

Since we know that
[tex]\lim_{x\rightarrrow 0}\frac{sin(x)}{x}= 1[/itex]<br /> I would be inclined, as a first step, to let y= 1/x. Then the limit becomes<br /> $$\lim_{y\rightarrow 0}\frac{sin(y)}{y^2(2/y- 1)}$$<br /> which, multiplying one of the y's from y² into the parentheses, gives<br /> [tex]\lim_{y\rightarrow 0}\frac{sin y}{y}\frac{1}{2-y}[/itex] <br /> which should be pretty easy.[/tex][/tex]

 

[statdad]

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

 

[Gib Z]

Since we know that
[tex]\lim_{x\rightarrrow 0}\frac{sin(x)}{x}= 1[/itex]<br /> I would be inclined, as a first step, to let y= 1/x. Then the limit becomes<br /> $$\lim_{y\rightarrow 0}\frac{sin(y)}{y^2(2/y- 1)}$$<br /> which, multiplying one of the y's from y² into the parentheses, gives<br /> [tex]\lim_{y\rightarrow 0}\frac{sin y}{y}\frac{1}{2-y}[/itex] <br /> which should be pretty easy.[/tex][/tex]

[tex]$$<br /> A very elegant solution Halls =]<br /> <br /> <blockquote data-attributes="" data-quote="statdad" data-source="post: 1899525" cite="https://www.physicsforums.com/goto/post?id=1899525" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-title"> statdad said: </div> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Just a question: does "x -> 00" mean the limit as x goes to zero or as x goes to infinity? </div> </div> </blockquote><br /> It means the limit as x goes to infinity, commonly used notation on forums for those who don't use latex.$$[/tex]

 

[farhad88]

derivation to funkton
f(x)= e^sin^22x when x=pi/2
i try but i get "0"... is this right solution?
please if anyone help me

 

[Mark44]

derivation to funkton
f(x)= e^sin^22x when x=pi/2
i try but i get "0"... is this right solution?
please if anyone help me

Please start a new thread rather than tacking your question onto a thread that is more than two years old.

 

[farhad88]

Please start a new thread rather than tacking your question onto a thread that is more than two years old.

com an man I am not so used by sympols in com puter .. but i well try

 

[farhad88]

calculate the derivative of the function. specify the exact answer

f(x)= e^sin^2(2x) when x=∏/2

 

[Mark44]

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

 

[farhad88]

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...