#lim_(x->0)(cosx-1)/x^2#
it's always worth plugging the actual limit in to see what you get. here it is #(cos0-1)/0^2 = (1-1)/0 = 0/0# ie indeterminate
multiplying top and bottom by #cos x + 1# gives us this
#lim_(x->0)((cosx-1)(cosx+1))/((cosx+1) x^2)#
#= lim_(x->0)(cos^2x-1)/((cosx+1) x^2)#
#= \color{red}{-} lim_(x->0)(sin^2x)/((cosx+1) x^2)#
#= - lim_(x->0)(sin^2x)/( x^2) 1/(cosx+1)#
#= - lim_(x->0)(sinx)/( x) lim_(x->0)(sinx)/( x) lim_(x->0) 1/(cosx+1)#
it is well known that #lim_(x->0)(sinx)/( x) = 1#
and we can conclude that # lim_(x->0) 1/(cosx+1) = 1/2# simply by plugging in #x = 0#
so the limit here is #- 1/2#
often, the the simplest approach with indeterminate forms is to use L'Hopital's rule. the problem is that sometimes you cannot use L'Hopital and this might be one of them. [using L'Hopital in connection with trig functions can be especially hazardous.]
trying to use L'Hopital illustrates the problem:
#lim_(x->0)(cosx-1)/x^2 = lim_(x->0)(- sin x)/(2x) = lim_(x->0)(- cos x)/2 = - 1/2#
here, the top and bottom were differentiated twice to get to that result and then 0 was plugged in to the determinate form
but you cannot use L'Hopital on # lim_(x->0)(- sin x)/(2x)# because that limit is used (and proved another way - Squeeze Theorem) in the derivation of the derivatives of #cos x# and #sin x#. so you would again have to stop there and rely again on the fact that it is well known that #lim_(x->0)(sinx)/( x) = 1#
i am not even that sure you can use it on the first step ie #lim_(x->0)(cosx-1)/x^2# because that equals #lim_(x->0 ) (\color{red}{(cosx-1)/x} * 1/x)#, so i will ask someone to double check the answer. that bit in red is also used in the derivation of the trig derivatives and proved in other ways....
the long and short is that i would recommend finding algebraic ways around these where possible. not only is it more interesting than always reaching for L'Hopital, but it is safer also.
