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

Question

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!

Emilia Aguilar · Accepted Answer

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

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#

Charles Arocha · Answer

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 terms cancel out on the numerator and denominator, so the denominator approaches 0. We don't need to consider the fact that the denominator is zero following direct substitution.

Ezekiel Alexander · Answer

undefined To show that $\lim_{x \to 0} \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{1}{2}$ using the Maclaurin series, we can express $\cos(x)$ and $\sin(x)$ as their respective Maclaurin series expansions:

$$
\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \ldots
$$
$$
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots
$$

Substitute these series expansions into the given expression and simplify:

$$
\frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{1 - (1 - \frac{x^4}{2!} + \frac{x^8}{4!} - \frac{x^{12}}{6!} + \ldots)}{x \cdot (x^3 - \frac{(x^3)^3}{3!} + \frac{(x^3)^5}{5!} - \frac{(x^3)^7}{7!} + \ldots)}
$$

Now, expand the numerator and denominator:

$$
\frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{\frac{x^4}{2!} - \frac{x^8}{4!} + \frac{x^{12}}{6!} - \ldots}{x^4 - \frac{x^{10}}{3!} + \frac{x^{16}}{5!} - \ldots}
$$

Cancel out the common factors of $x^4$:

$$
\frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{\frac{1}{2!} - \frac{x^4}{4!} + \frac{x^8}{6!} - \ldots}{1 - \frac{x^6}{3!} + \frac{x^{12}}{5!} - \ldots}
$$

Now, as $x$ approaches 0, terms with higher powers of $x$ become negligible, so the expression reduces to:

$$
\frac{1}{2} \cdot \frac{1}{1} = \frac{1}{2}
$$

Therefore, $\lim_{x \to 0} \frac{1 - \cos(x^2)}{x \sin(x^3)} = \frac{1}{2}$.

How can I show that #lim_(x-&gt;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!

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