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

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