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SpyIntel [72]
3 years ago
7

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

Mathematics
1 answer:
beks73 [17]3 years ago
6 0

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

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

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

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

and

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

So, we have

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

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