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

1. Oct 3, 2008

### 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'hopitals and i seem to be going nowhere

2. Oct 3, 2008

### jhicks

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

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

3. Oct 3, 2008

### cos(e)

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

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. Oct 3, 2008

### jhicks

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

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. Oct 3, 2008

### cos(e)

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

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

6. Oct 3, 2008

### jhicks

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

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. Oct 3, 2008

### Gib Z

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

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

8. Oct 3, 2008

### HallsofIvy

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

Since we know that
$$\lim_{x\rightarrrow 0}\frac{sin(x)}{x}= 1[/itex] I would be inclined, as a first step, to let y= 1/x. Then the limit becomes [tex]\lim_{y\rightarrow 0}\frac{sin(y)}{y^2(2/y- 1)}$$
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. Oct 3, 2008

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

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

10. Oct 3, 2008

### Gib Z

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

A very elegant solution Halls =]

It means the limit as x goes to infinity, commonly used notation on forums for those who don't use latex.

11. Feb 3, 2011

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

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. Feb 3, 2011

### Staff: Mentor

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

13. Feb 3, 2011

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

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

14. Feb 3, 2011

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

calculate the derivative of the function. specify the exact answer

f(x)= e^sin^2(2x) when x=∏/2

15. Feb 3, 2011

### Staff: Mentor

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