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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

8

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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