How do you evaluate #lim_(x->0)# #(cosax-cosbx)/x^2#?

1 Answer
Hammer
May 13, 2018

#lim_(x->0) (cosax-cosbx)/x^2 = (b^2-a^2)/2#

Explanation:

This is a calculus question and not a trigonometry one, even if it has the cosine function. Everything related to limits is calculus, so please, be more careful next time!

Now, getting past the wrong branch of mathematics, we see that our limit is a #0/0# situation:

#lim_(x->0) (cos0 - cos0)/0 = 0/0#

Thus, we can use L'Hôpital's rule:

#lim_(x->0) (cosax-cosbx)/x^2=lim_(x->0)("d"/("d"x)(cosax-cosbx))/("d"/("d"x)x^2)#

By the Chain rule, which states that

#[f(g(x))]' = f'(g(x))*g'(x)#

We get the following relations:

#"d"/("d"x) sinalphax=alphacosalphax#

#"d"/("d"x) cosalphax = -alphasinalphax#

for a constant #alpha#.

#=> lim_(x->0) (cosax-cosbx)/x^2=lim_(x->0)(-asinax+bsinbx)/(2x)#

Use L'Hôpital's rule again.

#=lim_(x->0) (-a^2cosax+b^2cosbx)/2#

Finally, plug in #x=0# to get the desired result:

#=(-a^2cos0 + b^2cos0)/2=color(red)((b^2-a^2)/2#

Related questions
Impact of this question
11820 views around the world
You can reuse this answer
Creative Commons License