Question

Lim x->pi/3 ((2cosx-1)/(tan^2x-3))

Answer 1

You can check that the limit comes in an undefined form:

`\displaystyle \lim_{x\to(\pi)/(3)} (2\cos(x)-1)/(\tan^2(x)-3)=(0)/(0)`

In these cases, we can use de l'Hospital rule, and evaluate the limit of the ratio of the derivatives. We have:

`\frac{\text{d}}{\text{d}x}2\cos(x)-1 = -2\sin(x)`

and

`\frac{\text{d}}{\text{d}x}\tan^2(x)-3 = 2(\tan(x))/(\cos^2(x))=(2\sin(x))/(\cos^3(x))`

So, we have

`\displaystyle \lim_{x\to(\pi)/(3)} (2\cos(x)-1)/(\tan^2(x)-3)=\lim_{x\to(\pi)/(3)} (-2\sin(x))/((2\sin(x))/(\cos^3(x)))=\lim_{x\to(\pi)/(3)}-\cos^3(x)=-\cos^3\left((\pi)/(3)\right)`