Evaluate #lim_(x->0)(1-cosx)/x^2#?

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Answer 1 · 2017-12-18T07:05:58

#1/2#

L'Hopital's rules says that the #lim_(x->a)(f(x))/(g(x))=>(f'(a))/(g'(a))#

Using this, we get #lim_(x->0)(1-cosx)/x^2=>(-sin0)/(2(0))#

Yet as the denominator is #0#, this is impossible. So we do a second limit:
#lim(x->0)(sinx)/(2x)=>(cos0)/2=1/2=0.5#

So, in total
#lim_(x->0)(1-cosx)/x^2=>lim_(x->0)(sinx)/(2x)=>cosx/2=>cos0/2=1/2#

Answer 2 · 2017-12-18T13:42:50

#lim_(xrarr0)(1-cosx)/x^2 = lim_(xrarr0)(sin^2x/x^2 * 1/(1+cosx)) = 1/2#

#(1-cosx)/x^2 = ((1-cosx))/x^2 * ((1+cosx))/((1+cosx))#

# = (1-cos^2x)/(x^2(1+cosx)#

# = sin^2x/(x^2(1+cosx)#

#= sin^2x/x^2 * 1/(1+cosx)#

Answer 3 · 2018-08-10T03:15:40

#1/2#

We see that through direct evaluation, we get indeterminate form, #0/0#. Next, we can use L'Hôpital's Rule, which says

#lim_(xtoa)(f(x))/g(x)=lim_(xtoa)(f'(x))/(g'(x))#

We can take the derivative of our numerator and denominator function to obtain the new limit

#lim_(xto0)(sinx)/(2x)#

Through direct evaluation, we would get indeterminate form again, so we can take the derivatives once more to get

#lim_(xto0)(cosx)/2#

When we evaluate this limit at zero, we get

#lim_(xto0)(1-cosx)/(x^2)=1/2#

Hope this helps!