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$$\quad\quad \lim_{ x \to 0} \frac {\sin 5 x } {\sin 2 x } $$

I don't know how to start, should I multiply by something... to simplify the expression or ...?

J. M. ain't a mathematician
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dramadeur
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7 Answers7

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As $x\to0,x\ne0$

So, divide the numerator & the denominator by $x$ to get

$$ \lim_{ x \to 0} \frac {\sin 5 x } {\sin 2 x }=\dfrac{5\lim_{x\to0}\dfrac{\sin5x}{5x}}{2\lim_{x\to0}\dfrac{\sin2x}{2x}}$$

lab bhattacharjee
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  • I thought you can't do that since it's like sin(5x), so dividing by 5x should get you $\frac {sin(5x)}{5x}$ no? – dramadeur May 04 '15 at 05:39
  • @dramadeur, That's what I've got. Not sure about "can't do that"? – lab bhattacharjee May 04 '15 at 05:41
  • But you're not dividing numerator and denominator by the same value. You divide numerator by 5x, whereas denominator by 2x... you can't do that for sure! Or is there a typo in your solution? Did you mean to divide by x? If so, sin(5x)/x doesn't equal to 1 – dramadeur May 04 '15 at 05:42
  • @dramadeur If you check the right side of the equality, you can easily verify that the expression is identical (algebraically) with the one on the left side, so what is the problem?? BTW, you can take away those two limits on the right to see this more clearly. – Timbuc May 04 '15 at 06:04
  • @dramadeur Sure, $5x$ and $2x$ are different, and that's why the answerer has added the multiplicative factor $\frac{5}{2}$. He multiplied by $\frac{5}{2}\frac{2x}{5x}=1$. – Guest May 04 '15 at 06:18
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Well, $$ \lim_{ x \to 0} \frac {\sin x } {x } = 1. $$ This implies following

$$ \lim_{ x \to 0} \frac { x } {\sin x } = 1, \quad \lim_{ x \to 0} \frac {\sin \alpha x } {x } = \alpha, \quad\text{and}\quad \lim_{ x \to 0} \frac { x } {\sin\beta x } = \frac{1}{\beta}. $$

In particular $$ \lim_{ x \to 0} \frac {\sin 5x } {x } = 5 \quad\text{ and }\quad \lim_{ x \to 0} \frac {x } {\sin 2x } = \frac{1}{2}. $$

Finally $$ \lim_{ x \to 0} \frac {\sin 5x } {\sin 2x } = \lim_{ x \to 0} \frac {x\sin 5x } {x\sin 2x } = \lim_{ x \to 0} \frac {\sin 5x } {x } \cdot \lim_{ x \to 0} \frac {x } {\sin 2x } = \frac{5}{2}. $$

In general you may always thinking that $\sin y \approx y$ when $y\approx 0$ ($\sin \alpha x \approx \alpha x$ if $x\approx 0$).

Nikita Evseev
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$$\begin{align} \displaystyle\lim_{x\to 0}\frac{\sin(5x)}{\sin(2x)}&=\lim_{x\to 0}\frac{\sin(5x)}{1}\cdot\frac{1}{\sin(2x)}\\ &=\lim_{x\to 0}\frac{\sin(5x)}{5x}\cdot\frac{2x}{\sin(2x)}\cdot\frac{5}{2}\\ &=\lim_{x\to 0}\frac{\sin(5x)}{5x}\cdot\lim_{x\to 0}\frac{2x}{\sin(2x)}\cdot\lim_{x\to 0}\frac{5}{2}\\ &=1\cdot1\cdot\frac{5}{2}\\ &=\frac{5}{2} \end{align}$$

  • I don't see how you replaced $\frac {sin(5x)}{1}$ by $\frac {sin(5x)}{5x}$ – dramadeur May 04 '15 at 05:47
  • @dramadeur $\displaystyle\lim_{x\to 0}\frac{\sin(x)}{x}=1$ . Please see this post: http://math.stackexchange.com/questions/75130/how-to-prove-that-lim-limits-x-to0-frac-sin-xx-1?lq=1 –  May 04 '15 at 05:48
  • yeah, but you don't have 1, you have $ \frac {sin(5x)}{1}$ – dramadeur May 04 '15 at 05:48
  • @dramadeur Look at at the next equality, you can see that the equality holds, but this time we are able to exploit the fact that $\displaystyle\lim_{x\to 0}\frac{\sin(x)}{x}=1$. –  May 04 '15 at 05:50
  • I don't see why would you need to write 1 as $\frac {sin(5x)}{5x}$ and then you got 5/2. It's like you skipped some steps, that could prove vital to my understanding of the solution. – dramadeur May 04 '15 at 05:54
  • @dramadeur I don't think I skipped any steps, you can verify each equality, and each of the fraction as x tends to 0. The $\frac{5}{2}$ is necessary to maintain the equality, as it will cancel out the $5x$ and the $2x$ we introduced. If you look at the 2nd equality, you can cancel everything out back to the original fraction you posed. Cheers. –  May 04 '15 at 05:57
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    It seems to be that the OP doesn't understand that $$\lim_{x\to0}\frac{\sin x}x=1\implies \lim_{x\to 0}\frac{\sin kx}{kx}=1;,;;k\neq 0$$ and, in fact, it is true that if we have a function s.t. $;f(x)\xrightarrow[x\to x_0]{}0;$ , then also $$\lim_{x\to x_0}\frac{\sin f(x)}{f(x)}=1$$ Without these basic facts most of the solutions here will be hardly understood. – Timbuc May 04 '15 at 06:07
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Some good answers here already, but L'Hopital can also be used in this case.

$$\lim_{x\to0}\frac{\sin(5x)}{\sin(2x)} =\lim_{x\to0}\frac{5\cos(5x)}{2\cos(2x)} = \frac{5\cdot1}{2\cdot1} = \underline{\underline{\frac52}}$$

Alec
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Use l'Hopital's rule when you appear to have zero divided by zero

David Quinn
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Using the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$

\begin{align} \lim_{x \to 0} \frac{\sin 5x}{\sin 2x} & = \lim_{x \to 0} \frac{\sin 5x}{\sin 2x} \frac{5x}{5x} \frac {2x}{2x} \\ & =\lim_{x \to 0} \frac{\sin 5x}{5x} \frac{2x}{\sin 2x} \frac{5x}{2x} \\ & = 1 \cdot 1 \cdot \frac52 \\ & = \frac 52\\ \end{align}

Wherein we exploited the fact that multiplying in fractions with the same numerator and denominator is the same thing as multiplying the whole thing by $1$.

mopy
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$\displaystyle\lim_{x\to0}\ \frac{sin(5x)}{sin(2x)}=\displaystyle\lim_{x\to0}\ \frac{5x-\frac{1}{3!}(5x)^3+\cdots}{2x-\frac{1}{3!}(2x)^3+\cdots}=\displaystyle\lim_{x\to0}\ \frac{5-\frac{5^3}{3!}x^2+\cdots}{2-\frac{2^3}{3!}x^2+\cdots}=\frac{5}{2}$

reluctant mathematician
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