Use series to evaluate the limit #\lim_(x\rarr0)(x^2/2-1-\cos(x))/x^4#?

Recall that #cosx = 1 - x^2/2 +x^4/(4!) + ...#

#L = lim_(x->0) (x^2/2 - 1 - (1 - x^2/2 + x^4/(4!) + ...))/x^4#

This clearly evaluates to nothing but #-oo#. However, if the question was #(x^2/2 - 1 + cosx)/x^4#, we would see that

#L = lim_(x->0) (x^2/2 - 1 + 1 - x^2/2 + x^4/(4!) + ...)/x^4#

#L = lim_(x->0) (x^4/(4!) - x^6/(6!) + x^8/(8!) + ....)/x^4#

#L = 1/(4!) + 0 + 0 + 0 + ...#

#L = 1/24#

Hopefully this helps!

Recall the Maclaurin Series expansion for cosine,

#cosx=sum_(n=0)^oo(-1)^nx^(2n)/((2n)!)=1-x^2/2+x^4/(4!)-x^2/(6!)+...#

We can insert this series into our limit (that is, the first few terms of the series, followed by the ellipsis to indicate that it goes on forever):

#lim_(x->0)(x^2/2-1-(1-x^2/2+x^4/(4!)-x^2/(6!)+...))/x^4#

And distribute the negative sign, allowing us to combine the relevant terms:

#lim_(x->0)(x^2/2-1-1+x^2/2-x^4/(4!)+x^6/(6!)-...)/x^4#

#lim_(x->0)((cancel2x^2)/cancel2-2-x^4/(4!)+x^6/(6!)-...)/x^4#

#lim_(x->0)(x^2-2-x^4/(4!)+x^6/(6!)-...)/x^4#

Divide all terms by #x^4:#

#lim_(x->0)1/x^2-2/x^4-1/(4!)+x^2/(6!)-...#

We see that #x^2/(6!)# vanishes as #x->0,# and so would all terms that come after it, as all the following terms would have an #x# raised to increasingly higher powers in the numerator.

So, since all of the subsequent terms vanish, we won't include ellipsis anymore as we're no longer dealing with them.

Perform some simplification on the first two terms to take the limit:

#1/x^2-2/x^4=(x^4-2x^2)/x^6=(x^2(x^2-2))/x^6#

#=(x^2-2)/(x^4)#

Take the limit, and note that the constant #-1/(4!)# ends up having no impact due to the infinite nature of the limit:

#lim_(x->0)(x^2-2)/(x^4)-1/(4!)=(-2)/0-1/(4!)=-oo#