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

Thanks!

3 Answers

1/2

Explanation:

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

Dec 18, 2017

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

Explanation:

(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)

Aug 10, 2018

1/2

Explanation:

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!