Question

How can I show that `lim_(x->0)(1-cos(x^2))/(xsin(x^3))=1/2` using the Maclaurin series?

Show that `lim_(x->0)(1-cos(x^2))/(xsin(x^3))=1/2` using the Maclaurin series. I know how maclaurin series works but that's ridiculous!

Answer 1

`lim_(x->0) (1-cosx^2)/(xsin^3x) = 1/2`

Explanation:

Consider the MacLaurin series for `sint` and `cost`:

`sint = sum_(n=0)^oo (-1)^n t^(2n+1)/((2n+1)!)`

`cost = sum_(n=0)^oo (-1)^n t^(2n)/((2n)!)`

Substituting `t=x^3` in the first and `t=x^2`in the second, we have:

`sinx^3 = sum_(n=0)^oo (-1)^n x^(6n+3)/((2n+1)!)`

`cosx^2 = sum_(n=0)^oo (-1)^n x^(4n)/((2n)!)`

Stop the expansion of `cosx^2` at the second term to have:

`1-cosx^2 = 1 - (1-x^4/2+o(x^8)) = x^4/2+o(x^8)`

and:

`(1-cosx^2)/x = x^3/2+o(x^7)`

Now use only the first term of the expansion of `sinx^3`:

`sinx^3 = x^3-o(x^9) = x^3(1-o(x^6))`

so that:

`(1)/(sinx^3) = 1/(x^3(1-o(x^6))`

Now consider the geometric series:

`sum_(n=0)^oo x^n =1/(1-x)`

so that:

`1/(1-x) = 1+x+o(x^2)`

and then:

`1/(1-o(x^6)) = 1+o(x^6) +o(o(x^6)^2)= 1+o(x^6)`

`1/(sinx^3) = 1/x^3(1+o(x^6)) = 1/x^3 + o(x^3)`

Finally:

`(1-cosx^2)/(xsin^3x) = ( x^3/2+o(x^7) ) (1/x^3 + o(x^3) )`

`(1-cosx^2)/(xsin^3x) = 1/2 + x^3/2o(x^3)+(o(x^7))/x^3 + o(x^6)o(x^3 )`

Now by definition of `o(x^n)` we have that:

`lim_(x->0) x^3(o(x^3)) = 0`

`lim_(x->0) (o(x^7))/x^3 = 0`

`lim_(x->0) o(x^6)o(x^3)= 0`

and then:

`lim_(x->0) (1-cosx^2)/(xsin^3x) = 1/2`

Answer 2

Numerator:
First write `cos(x^2)` as a MacLaurin series by doing a substitution with the MacLaurin series representation for
`cosu = sum_(n=0)^(oo) frac{(-1)^n u^(2n)}{(2n)!}`

So:
`cos(x^2) = sum_(n=0)^(oo) frac{(-1)^n x^(4n)}{(2n)!}`

`= 1 - frac{x^4}{2!} + frac{x^8}{4!} - (x^12)/(6!) + ... + frac{(-1)^nx^(4n)}{(2n)!}`

`1 - cos(x^2) = frac{x^4}{2!} - (x^8)/(4!) + (x^12)/(6!)- ... + frac{(-1)^nx^(4n)}{(2n)!}`

Denominator
Also use the MacLaurin series representation for `sin(u)` to substitute:
`sin(u) = sum_(n=0)^(oo) frac{(-1)^n u^(2n+1)}{(2n+1)!}`

`sin(x^3) = sum_(n=0)^(oo) frac{(-1)^n x^(6n+3)}{(2n+1)!}`

`xsin(x^3) = sum_(n=0)^(oo) frac{(-1)^n x^(6n+4)}{(2n+1)!}`

Expanded:
`= x^4 - frac{x^10}{3!} + (x^16)/(5!) - ... + frac{(-1)^n x^(6n+4)}{(2n+1)!}`

Limit:
`lim_(x->0) frac{(frac{x^4}{2} - (x^8)/(4!) + (x^12)/(6!)- ... + frac{(-1)^nx^(4n)}{(2n)!})}{(x^4 - frac{x^10}{3!} + (x^16)/(5!) - ... + frac{(-1)^n x^(6n+4)}{(2n+1)!})}`

By direct substitution, the numerator approaches `1/2` (because `x^4` is the only term that evenly cancels out on the numerator and denominator)

The denominator approaches 0. We don't have to take into account the fact that the denominator is zero after direct substitution because the terms cancel out on the numerator and denominator.