Trig Limit Homework Help: Solving (2sinxcosx)/ (2x^2 + x) at x=0

[jog511]

Homework Statement

lim x->0 (2sinxcosx)/ (2x^2 + x )

Homework Equations

2sinxcosx = sin(2x)

The Attempt at a Solution

denom. factors to x(2x +1) how to proceed?

 

[Mark44]

Homework Statement

lim x->0 (2sinxcosx)/ (2x^2 + x )

Homework Equations

2sinxcosx = sin(2x)

The Attempt at a Solution

denom. factors to x(2x +1) how to proceed?

Your expression can be rewritten as $\frac{\sin(2x)}{2x(x + 1/2)}$. Does that help?

 

[jog511]

the limit of sin2x/2x will be 2. am I on the right track?

 

[jog511]

sin2x/2x will be 1 I mean

 

[jog511]

can it be written as sin2x/2x * 1/(x + 1/2)

 

[PeroK]

can it be written as sin2x/2x * 1/(x + 1/2)

Yes, that's the idea.

 

[jog511]

I get lim = 2

 

[jog511]

very helpful

 

[PeroK]

I get lim = 2

Yes, but note that you could actually have done it directly:

$$\lim_{x→0}\frac{2sin(x)cos(x)}{2x^2+x}= 2\lim_{x→0}\frac{sin(x)}{x}cos(x)\frac{1}{2x+1}=2$$