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I need help evaluating the following limit without using l'Hopital's rule:

$$\lim_{x\to \pi/3} \frac{\sin{x}-\sqrt{3}\cos{x}}{\sin{(x-\pi/3)}}$$

I have tried converting $\sin{(x-\pi/3)}$ to $-\cos{(\pi/6+x)}$ and seeing if I can cancel out terms. I have also tried rationalizing the expression by multiplying both numerator and denominator with $(\sin{x} + \sqrt{3}\cos{x})$. Neither seemed to lead me where I needed and I am out of ideas

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    $\begingroup$ what have you tried? $\endgroup$
    – IITM
    May 31, 2021 at 0:59
  • $\begingroup$ I have tried converting sin (x-pi/3) to -cos(pi/6+x) and seeing if I can cancel out terms. I have also tried rationalizing the expression by multiplying both numerator and denominator with (sinx + v3cosx). Neither seemed to lead me where I needed and I am out of ideas. $\endgroup$ May 31, 2021 at 1:03
  • $\begingroup$ It would be great if you can add this in the question,otherwise the question might attract downvotes,close votes $\endgroup$
    – IITM
    May 31, 2021 at 1:08
  • $\begingroup$ Thanks for the tip! I'll add it $\endgroup$ May 31, 2021 at 1:09
  • $\begingroup$ Even faster. Let $x=y+\frac \pi 3$ and expand the numerator $\endgroup$ May 31, 2021 at 5:37

1 Answer 1

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$$\lim_{x\to \pi/3} \frac{\sin{x}-\sqrt{3}\cos{x}}{\sin{(x-\pi/3)}}$$

note that $$\sin \left(x-\frac{\pi}{3}\right) = \sin x \cos \frac{\pi}{3}-\cos x \sin \frac{\pi}{3}$$

$$\sin \left(x-\frac{\pi}{3}\right)= \frac{1}{2}\sin x-\frac{\sqrt{3}}{2} \cos x$$

substitute this in denominator,to get the answer

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  • $\begingroup$ Thank you! This solves it. I was not aware of this conversion, but it comes in handy! $\endgroup$ May 31, 2021 at 1:19

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