Question

Lim x->0 √1-cos 2x/√2x=? Thanks

Answer 1

`lim_(x->0) (sqrt(1-cos(2x))/sqrt(2x)) = 0`

Explanation:

Disclaimer: I don't know much about limits besides the general idea of what they are, but here's my take on this:

Looking at the limit:
`lim_(x->0) (sqrt(1-cos(2x))/sqrt(2x))`
`iff sqrt(lim_(x->0) ((1-cos(2x))/(2x))`

Using l'hopital's method, we can see that both the numerator and denominator is approaching zero, so:
If
`lim_(x->c) (f(x)) = lim_(x->c) (g(x)) = 0 or +-oo`

then `lim_(x->c) (f(x))/g(x) = lim_(x->c) (f'(x))/(g'(x))`

In this case `f(x) = 1-cos(2x)`, `f'(x) = 2sin(2x)`
`g(x) = 2x`, `g'(x) = 2`

Therefore:
`sqrt(lim_(x->0) ((1-cos(2x))/(2x))` `= sqrt(lim_(x->0) ((2sin(2x))/2)`

Now we have eliminated the problem of dividing by zero, so
`sqrt(lim_(x->0) ((2sin(2x))/2)` `= sqrt(0/2) = 0`

Answer 2

`lim_(xrarr0)sqrt(1-cos2x)/sqrt(2x)=sqrt(lim_(xrarr0)(1-cos2x)/(2x))`

Knowing that the Maclaurin series for `cosx` is `cosx=sum_(n=0)^oo((-1)^nx^(2n))/((2n)!)=1-x^2/(2!)+x^4/(4!)-x^6/(6!)+...`, we can see that `cos2x=sum_(n=0)^oo((-1)^n(2x)^(2n))/((2n)!)=1-(4x^2)/(2!)+(16x^4)/(4!)-(64x^6)/(6!)+...`

Then, we can say the limit is:

`=sqrt(lim_(xrarr0)(1-(1-(4x^2)/(2!)+(16x^4)/(4!)-(64x^6)/(6!)+...))/(2x))`

`=sqrt(lim_(xrarr0)((4x^2)/(2!)-(16x^4)/(4!)+(64x^6)/(6!)+...)/(2x))`

`=sqrt(lim_(xrarr0)(2x)/(2!)-(8x^3)/(4!)+(32x^6)/(6!)+...)`

As the series continues infinitely, all the terms contain an `x` and will become `0` as `xrarr0`.

`=sqrt0`

`=0`

Answer 3

`lim_(x->0)sqrt(1-cos(2x))/sqrt(2x) = 0`

Explanation:

`lim_(x->0)sqrt(1-cos(2x))/sqrt(2x) = lim_(x->0)sqrt((1-cos(2x))/(2x))`

Using the substitution `theta = 2x`, when `x` goes to `0`, `theta` also goes to `0`

`lim_(x->0)sqrt((1-cos(theta))/(theta))`

Which, if you look closely, you'll see is the square root of a fundamental limit, one we know to be 0. So the limit as a whole is `0`

(If you don't quite recognize it as a fundamental limit, you can tweak it so it involves the other fundamental limit `lim_(x->0)sin(x)/x = 1`. A non-circular proof of that, however, requires either to use the epsilon-delta definition of limits or some geometric arguments)