Question

What is the limit of `(cos(3x)-cos(4x))/x^2` as x approaches `0`?

Answer 1

the answer
`lim_(xrarr0)[16cos(4x)-9co(3x)]/2=[16-9]/2=7/2`

Explanation:

show below

`lim_(xrarr0)(cos(3x)-cos(4x))/x^2`

Direct compensation product equal `0/0` so we will use l'hospital rule

`lim_(xrarra)f(x)/g(x)=0/0`

`lim_(xrarra)(f'(x))/(g'(x))`

`f(x)=cos(3x)-cos(4x)`

`f'(x)=4sin(4x)-3sin(3x)`

`g(x)=x^2`

`g'(x)=2x`

`lim_(xrarr0)[4sin(4x)-3sin(3x)]/(2x)`

since the Direct compensation product again equal `0/0` so we will derive it again

`f''(x)=16cos(4x)-9co(3x)`

`g''(x)=2`

`lim_(xrarr0)[16cos(4x)-9co(3x)]/2=[16-9]/2=7/2`