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Find $$ \lim_{x\to \frac{\pi}{2}} \frac{tan(3x)}{tan(7x)} $$

I want to find it using l'hopital's rule

My answer was :

$$ \frac{\frac{sin(3x)}{cos(3x)}}{\frac{sin(7x)}{cos(7x)}} $$

$$ \frac{\frac{cos(3x) \cdot 3 \cdot cos(3x) - sin(3x) \cdot (-sin3x \cdot 3)}{cos^23x}}{\frac{cos7x \cdot 7 \cdot cos7x - sin7x \cdot (-sin7x \cdot 7 )}{cos^27x}} $$

$$ \frac{\frac{27cos^2x - 27sin^2x}{3cos^2x}}{\frac{343cos^2x - 343sin^2x}{7cos^2x}} $$

I don't know guys i got confused. am i right until this step ? should i use l'hopital's rule again ? or there is something wrong with my solution ?

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  • $\begingroup$ As the answer shows, knowing the derivative of $\tan$ is not necessary to do this problem. However, it is important enough to be worth memorizing. if you had known it and applied it here, the problem would have been pretty easy. By the way, trig functions look nicer with \ in front: $\sin$ is obtained by "\sin" $\endgroup$
    – Stefan Smith
    Nov 13, 2013 at 0:29

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$$\frac{\tan{3x}}{\tan{7x}} = \frac{\sin{3x}\cos{7x}}{\sin{7x}\cos{3x}}$$ Notice that $$\lim_{x\to \frac{\pi}{2}} \frac{sin(3x)}{sin(7x)} = \frac{sin(\frac{3\pi}{2})}{sin(\frac{7\pi}{2})} = \frac{-1}{-1} = 1$$ and $$\lim_{x\to \frac{\pi}{2}} \frac{cos(7x)}{cos(3x)}\stackrel{L'Hopital}{=} \lim_{x\to \frac{\pi}{2}} \frac{7}{3}\frac{sin(7x)}{sin(3x)} = \frac{7}{3}$$

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  • $\begingroup$ Quicker than my answer, so I took mine out. $\endgroup$
    – imranfat
    Nov 12, 2013 at 22:22
  • $\begingroup$ sorry but i don't know if this is a dumb question, but how would i know that $ sin(\frac{7\pi}{2}) = -1 $ $\endgroup$
    – Out Of Bounds
    Nov 13, 2013 at 0:40
  • $\begingroup$ @Tennisman Since sin is an odd and a periodic function with a period $2\pi$ then we can say that $\sin{\frac{7\pi}{2}} = \sin (\frac{7\pi}{2}-2\pi) =\sin ( -\frac{\pi}{2}) = -\sin(\frac{\pi}{2}) = -1$ $\endgroup$
    – sve
    Nov 13, 2013 at 14:43

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