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I want to find $$\lim \limits_{x\to 0}{\sin{42x} \over \sin{6x}-\sin{7x}}$$ without resorting to L'Hôpital's rule. Numerically, this computes as $-42$. My idea is to examine two cases: $x>0$ and $x<0$ and use ${\sin{42x} \over \sin{6x}}\to 7$ and ${\sin{42x} \over \sin{7x}}\to 6$. I can't find the appropriate inequalities to use the squeeze theorem, though. Do you have suggestions?

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    $\begingroup$ taylor's always the play with these in my opinion $\endgroup$ – qbert Jan 16 '17 at 19:57
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Hint:

$$\frac{\sin42x}{\sin6x-\sin7x}=\cfrac{\frac{\sin42x}{42x}}{\frac17\frac{\sin6x}{6x}-\frac16\frac{\sin7x}{7x}}$$

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I provide another approach which uses the simpler limit $\lim\limits_{x \to 0}\cos x = 1$ compared to $\lim\limits_{x \to 0}\dfrac{\sin x}{x} = 1$. Let $x = 2t$ and the given expression can be rewritten as $$-\frac{\sin 84t}{2\cos 13t\sin t}$$ and $\cos 13t \to 1$ therefore the desired limit is equal to limit of $-\dfrac{\sin 84t}{2\sin t}$ as $ t\to 0$. Now it is easy to prove via induction that $$\lim_{t\to 0}\frac{\sin nt}{\sin t} = n$$ for all positive integers $n$ and therefore the desired limit is $-84/2 = -42$.


On request of user "Simple Art" (via comments) I show via induction that $$\lim_{t \to 0}\frac{\sin nt}{\sin t} = n\tag{1}$$ for all positive integers $n$. In what follows I will use the result that $\cos t \to 1$ as $t \to 0$ (and nothing more than that).

For $n = 1$ we see that the claim holds. Let's suppose that it holds for $n = m$ so that $(\sin mt)/\sin t \to m$ as $t \to 0$. Now we can see that $$\frac{\sin (m + 1)t}{\sin t} = \frac{\sin mt}{\sin t}\cos t + \cos mt$$ and letting $t \to 0$ we see that $$\lim_{t \to 0}\frac{\sin(m + 1)t}{\sin t} = m \cdot 1 + 1 = m + 1$$ so that the claim holds for $n = m + 1$. Thus $(1)$ holds for all positive integers $n$. It is easy to extend the claim for all rational values of $n$.


The whole point of the above gymnastics (as compared to the simpler and beautiful hint by "Simple Art") is to show that the current question can be solved by using a simpler limit $\cos t \to 1$ as $t \to 0$ instead of using the slightly more complicated limit $(\sin t)/t \to 1$ as $t \to 0$.


Update: Once the limit formula $$\lim_{x\to 0}\frac{\sin nx} {\sin x} =n$$ is available for positive integer $n$, the current question is easily solved by dividing the numerator and denominator by $\sin x$ and then taking limit to get $42/(6-7)=-42$ as answer. I wonder why I converted the difference in denominator to a product. Perhaps achieving simplicity is not simple.

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  • $\begingroup$ Haha, why prove by induction? $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 12:59
  • $\begingroup$ @SimpleArt: because I don't want to use $(\sin x) /x\to 1$ and just want to use $\cos x\to 1$. Remember that $\cos x\to 1$ deals with continuity of circular functions whereas $(\sin x) /x \to 1$ deals with differentiability of circular functions and hence is at a higher level in terms of complexity. $\endgroup$ – Paramanand Singh Jan 17 '17 at 14:33
  • $\begingroup$ Hm, could you expand on the inductive step for me? $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 14:40
  • $\begingroup$ @SimpleArt: see update to my answer. $\endgroup$ – Paramanand Singh Jan 17 '17 at 14:47
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    $\begingroup$ @SimpleArt: I also added some remarks about your answer. I hope you will like it. See the last paragraph in my answer. $\endgroup$ – Paramanand Singh Jan 17 '17 at 14:50

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