Evaluate #lim_(x->0)(1-cosx)/x^2#?
Thanks!
Thanks!
3 Answers
Explanation:
L'Hopital's rules says that the
Using this, we get
Yet as the denominator is
So, in total
Explanation:
# = (1-cos^2x)/(x^2(1+cosx)#
# = sin^2x/(x^2(1+cosx)#
#= sin^2x/x^2 * 1/(1+cosx)#
Explanation:
We see that through direct evaluation, we get indeterminate form,
We can take the derivative of our numerator and denominator function to obtain the new limit
Through direct evaluation, we would get indeterminate form again, so we can take the derivatives once more to get
When we evaluate this limit at zero, we get
Hope this helps!