Question

Find the limit. (Special Trig functions)
As the lim (sinx/2x^2-x) approaches 0.

Answer 1

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!

Answer 2

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

Answer 3

\[lim_{x\rightarrow 0} \frac{sin \ x}{2x^2 - x} = lim_{x\rightarrow 0} \frac{\frac{sin \ x}{x}}{2x - 1} \]

plus \[lim_{x\rightarrow 0} \frac{sin}{ x} = 1\]

Answer 4

Still having trouble?

Answer 5

yeah I still don't understand..

Answer 6

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

Answer 7

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

Answer 8

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

Answer 9

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

Answer 10

this the important bit @karatechopper !

https://gyazo.com/ad5c26d3cf8e2b41c20a8c7935b34a83

splitting it out so you have a \(\frac{sinx}{x}\) term

Answer 11

@ChillOut hi

Answer 12

understood! thanks :)

Answer 13

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

Answer 14

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

Answer 15

@karatechopper