Lim x->0 √1-cos 2x/√2x=? Thanks

3 Answers
Jun 27, 2017

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

Jun 28, 2017

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

Jun 28, 2017

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)