## karatechopper one year ago Find the limit. (Special Trig functions) As the lim (sinx/2x^2-x) approaches 0.

1. karatechopper

I broke the limit down and did this: sinx sinx ---- - ----- x 2x-1 I used the special trig function to find out that sinx/x=1. So now I am left with 1 - sinx/2x-1 And I'm stuck!

2. karatechopper

to make more sense- I am stuck on 1- (sinx/2x-1)

3. IrishBoy123

$lim_{x\rightarrow 0} \frac{sin \ x}{2x^2 - x} = lim_{x\rightarrow 0} \frac{\frac{sin \ x}{x}}{2x - 1}$ plus $lim_{x\rightarrow 0} \frac{sin}{ x} = 1$

4. ChillOut

Still having trouble?

5. karatechopper

yeah I still don't understand..

6. IrishBoy123

$$\large \frac{1}{2(0) - 1} = ???$$

7. ChillOut

You need to notice that 2x²-x = x(2x-1). That's what @IrishBoy123 did.

8. ChillOut

And that's the same as $$\frac{ \sin x }{ x }*\frac{ 1 }{ 2x-1 }$$

9. IrishBoy123

the crucial thing here is that $lim_{x\rightarrow 0} \frac{sin \ x}{x} = 1$ so you set it up to have a $$\frac{sin \ x}{x}$$ term but that $$\frac{sin \ x}{x}$$ term is seminal

10. IrishBoy123

this the important bit @karatechopper ! https://gyazo.com/ad5c26d3cf8e2b41c20a8c7935b34a83 splitting it out so you have a $$\frac{sinx}{x}$$ term

11. anonymous

@ChillOut hi

12. karatechopper

understood! thanks :)

13. karatechopper

BUT. How can you just take that x to the numerator when its being multiplied in the denominator?

14. ChillOut

When you have something like $$\LARGE\frac{\frac{a}{b}}{c}$$ you can rewrite that as $$\LARGE\frac{a}{b*c}$$.

15. ChillOut

@karatechopper