0

Does anyone have any idea how to transform this to calculate the limit as x approaches $\frac {\pi}{6}$, without using L'Hospital rule?

$$ \lim_{x\to \Large{\frac {\pi}{6}}} \left(\frac {\sin(x- \frac{\pi}{6})}{\frac{{\sqrt3}}{2}-\cos x}\right)$$

Blue
  • 83,939
kontojulii
  • 9
  • 1

3 Answers3

2

Let put $t=x-\frac{\pi}{6}$

we will compute

$$\lim_{t\to 0}\frac{\sin(t)}{\frac{\sqrt{3}}{2}-\cos(t+\frac{\pi}{6})}=$$

$$\lim_{t\to0}\frac{\sin(t)}{ \frac{\sqrt{3}}{2}(1- \cos(t))+\frac{\sin(t)}{2} }=$$

$$\lim_{t\to 0}\frac{1}{ \frac{1}{2}+\frac{\sqrt{3}}{2}\frac{t^2(1-\cos(t))}{t^2\sin(t)} }=2.$$

hamam_Abdallah
  • 63,719
0

Let $f(x) = \sin (x-\pi/6), g(x) = \cos x.$ The expression takes the form

$$-\frac{f(x) - f(\pi/6)}{g(x) - g(\pi/6)} = -\frac{(f(x) - f(\pi/6))/( x-\pi/6)}{(g(x) - g(\pi/6))/(x-\pi/6)}.$$

As $x\to \pi/6,$ the last expression $\to -f'(\pi/6)/g'(\pi/6)$ by the defintion of the derivative. That is easy to calculate, and no L'Hopital was used.

zhw.
  • 107,943
  • Lmao, you did not just pull that – Simply Beautiful Art Nov 10 '16 at 21:26
  • I don't understand your comment. – zhw. Nov 10 '16 at 22:07
  • You basically use L'Hospital's rule, which the OP explicitly said they did not want. – Simply Beautiful Art Nov 10 '16 at 23:01
  • Please point out the step where I used L'Hopital. – zhw. Nov 11 '16 at 00:48
  • Your entire answer is L'Hospital's rule. If one is told not to use $\frac d{dx}x^n=nx^{n-1}$ for $\frac d{dx}\sqrt x$, one should not follow straight through the proof of the power rule for the particular problem. The context implies you should follow a different path, like rationalizing or some other manner. – Simply Beautiful Art Nov 11 '16 at 00:56
  • 1
    But one is told to not use L'Hopital here, and I didn't. Sorry you cannot understand this elementary point. Everything you've said in these comments is wrong, and I have to say you're pretty rude about it. – zhw. Nov 11 '16 at 16:42
  • I'm sorry if I sound rude. My point is that L'Hospital's rule is more than just taking the derivative of the numerator and denominator, and that what L'Hospital's rule is actually trying to tell us is what you did above. So in essence, you applied L'Hospital's rule. If you are still confused, refer to the link in my first comment. – Simply Beautiful Art Nov 11 '16 at 17:04
  • L'Hopital's Rule is a theorem. I didn't use that theorem - at all. The difference between what I did and L'Hopital is important. Let me illustrate with an exercise: Assume $f(0)=g(0)=0,$ and $f'(0) = 4, g'(0)=3.$ What is

    $$\tag 1 \lim_{x\to 0}\frac{f(x)}{g(x)}?$$

    I would do this just I did the OP's problem. Try using L'Hopital here. Forget about it; you can't. Thus there has to be a difference in the approaches. Continued below …

     – zhw. Nov 11 '16 at 18:46
  • … Even if L'Hopital is allowed, I always first try the other approach if feasible. L'Hopital is like pushing a button in a video game. The other approach is fundamental and basic, often as easy, and sometimes easier. And with it the student gets to connect (and reconnect) with one of the most important concepts in mathematics: the definition of the derivative. (Believe me, they need this.) I'll stop here, having ranted elsewhere on this topic.

    http://math.stackexchange.com/questions/1286699/whats-wrong-with-lhopitals-rule/1286806#1286806

     – zhw. Nov 11 '16 at 18:47
  • Well, for L'Hospitals to be applied above, one needs more information. And I just wanted to be sure if you understood what you were doing here. – Simply Beautiful Art Nov 12 '16 at 00:38
0

Using trig identities $$\sin 2\theta = 2 \sin\theta \cos\theta\qquad\text{and}\qquad \cos\alpha - \cos\beta = -2 \sin\frac{\alpha - \beta}{2}\sin\frac{\alpha + \beta}{2}$$ ... we can write ...

$$\begin{align} \sin\left( x - \frac{\pi}{6}\right) &= 2\sin\left(\frac{x}{2} - \frac{\pi}{12}\right) \cos\left(\frac{x}{2} - \frac{\pi}{12}\right) \tag{1} \\[8pt] \frac{\sqrt{3}}{2}-\cos x &= \cos\frac{\pi}{6} - \cos x \\[4pt] &= 2\sin\left(\frac{\pi}{12} - \frac{x}{2}\right) \sin\left(\frac{\pi}{12} + \frac{x}{2}\right) \\[8pt] &= -2\sin\left(\frac{x}{2} - \frac{\pi}{12}\right) \sin\left(\frac{x}{2} + \frac{\pi}{12}\right) \tag{2} \end{align}$$

... so that, for $\sin\left(\frac{x}{2} -\frac{\pi}{12}\right) \neq 0$ (that is, $x$ in a neighborhood of $\frac{\pi}{6}$ ), ...

$$\frac{\sin\left( x - \dfrac{\pi}{6}\right)}{\dfrac{\sqrt{3}}{2} - \cos x} = - \frac{\cos\left(\dfrac{x}{2}-\dfrac{\pi}{12}\right)}{\sin\left(\dfrac{x}{2} + \dfrac{\pi}{12}\right)} \tag{3}$$

Taking the limit on the right-hand side of $(3)$ amounts to substitution.

Blue
  • 83,939