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Is this the correct way to solve for the limit?

I found the limit $x\to 0$ of $\sin4x/ \tan7x$ by these steps

1) $\dfrac{\sin4x}{1} \cdot \dfrac{\cos7x}{\sin7x}$

2) I crossed out the $\sin x$ in the numerator and denominator leaving me with $\dfrac{4\cos7x}{7}$

3) $4 (\cos7(0)) = 4 (\cos0=1)$

4) I was left with $ 4/7$

amWhy
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user44816
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    This would be a good response for this post. – David Mitra Dec 18 '12 at 16:19
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    No, your reasoning to get to 2) is not correct. Given $\sin (4x)\over \sin(7x)$ you cannot "cross out the $\sin x$" to obtain ${\sin (4x)\over \sin(7x)}={4\over 7}$. This is not true... – David Mitra Dec 18 '12 at 16:23
  • If this is not the correct method, how would you solve this example @David Mitra – user44816 Dec 18 '12 at 16:47
  • As in the answers below. What would be correct is to say is that $\lim_{x\rightarrow0}{\sin(4x)\over\sin(7x)}={4\over7}$. To prove this, you'd use the fact (which you should have encountered if this problem is posed to you) that $\lim_{x\rightarrow0}{\sin(x)\over x}={1}$:$$\lim_{x\rightarrow0}{\sin(4x)\over\sin(7x)}=\lim_{x\rightarrow0}\Bigl[4\cdot{\sin(4x)\over 4x}\cdot {1\over7}\cdot{ 7x\over \cdot\sin( 7x)}\Bigr]=4\cdot 1\cdot{1\over7}\cdot1={4\over7}.$$ Then you can use this to find your limit. – David Mitra Dec 18 '12 at 16:59
  • @David Mitra Where are you getting the x in the denominator (4x) and numerator (7x) from? – user44816 Dec 18 '12 at 17:16
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    Those guys and the constant factors $4$ and ${1\over7}$ do cancel (since $x\ne0$); so it's ok to just put them there. They were put there because, then, one can compute the limits $\lim_{x\rightarrow0}{\sin(4x)\over 4x}$ and $\lim_{x\rightarrow0}{7x\over \sin(7x)}$. – David Mitra Dec 18 '12 at 17:20
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    @user44816 Basically, what is being done is that we want to get something of the form $\frac{\sin y}{y}$. Since we have $\sin 4x$, let $y = 4x$. Then $\frac{\sin y}{y} = \frac{\sin 4x}{4x}$. Finally, since $y \to 0$ as $x \to 0$, we don't have to change the target of our limit. Now we have a nice situation where we can use $\lim_{y \to 0} \frac{\sin y}{y} = 1$. – Emily Dec 18 '12 at 17:38

3 Answers3

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$$\lim_{x\to 0}\frac{\sin 4x}{\tan 7x}=\frac47\cdot\frac{\lim_{x\to 0}\frac{\sin 4x}{4x}}{\lim_{x\to 0}\frac{\sin7x}{7x}}\cdot\lim_{x\to 0} \cos7x=\frac47\frac11\cdot1=\frac47$$

lab bhattacharjee
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Note that$$\frac{\sin 4x}{\tan 7x}=\frac{\sin 4x}{\sin 7x}\cos 7x=\frac{\sin 4x}{x}\frac{x}{\sin 7x}\cos 7x=\frac{\sin 4x}{x}\frac{1}{\frac{\sin 7x}{x}}\cos 7x$$ and so $$\lim_{x\to 0}\frac{\sin 4x}{\tan 7x}=4\frac{1}{7}$$ Your reasoning is not correct, one can't simply "cross out" $\sin 4x$ with $\sin 7x$

Nameless
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So, $\displaystyle \frac{\sin 4x}{\sin 7x} = \frac{4}{7}$ ? This uses one of those useful "facts" of magic triginometry. It makes manipulation of trig functions so easy that regardless of the fact that it is completely untrue one will not abandon its use.

So, $\sin 4x = 4 \sin x$ ? Well, if that's true then if we let $x = \pi/2$ we get

$$\sin 4 (\frac{\pi}{2}) = 4 \sin (\frac{\pi}{2}) = 4 \cdot 1 = 4$$

which has already led to a "surprising" result. I was under the impression that

$$-1 \leq \sin x \leq 1$$

But this flawless use of a basic magical trig fact has led me to realize that $\sin x$ can have any value I wish. And there's a bonus fact!! Since

$$\sin 4(\pi/2) = \sin \frac{4\pi}{2} = \sin 2\pi = 0$$,

if we string this true result together with the magic result previously obtained we get a common conclusion based on magic math, that is

$$0 = \sin 2\pi = \sin \frac{4\pi}{2} = \sin 4(\pi/2) = 4 \sin (\frac{\pi}{2}) = 4 \cdot 1 = 4$$

The magic "fact" comes from problems like this (and extremely wishful thinking)

$$\lim_{x \longrightarrow 0} \frac{\sin 4x}{x} = 4$$

Clearly, this must be the result of the "fact" that $\sin 4x = 4 \sin x$ ! Right ?

Not so fast Grasshopper. Here's what actually happened . . .

$$\lim_{x \longrightarrow 0}\frac{\sin 4x}{x} = \lim_{x \longrightarrow 0}\frac{4}{4} \cdot \frac{\sin 4x}{x} = 4 \lim_{x \longrightarrow 0}\frac{\sin 4x}{4x} = 4 \cdot 1 = 4$$

This is true because

$$\lim_{x \longrightarrow 0}\frac{\sin a x}{ ax} = 1 \hspace{.1in}\mbox{ where } \hspace{.1in} a \in \mathbb{R}$$

Using this fact and some legal algebra we can get the correct answer . . .

$$\lim_{x \longrightarrow 0}\frac{\sin 4 x}{ \tan 7x} = \lim_{x \longrightarrow 0}\frac{\sin 4 x}{ \left(\frac{\sin 7x}{\cos 7x}\right)} = \lim_{x \longrightarrow 0}\left[\frac{\sin 4 x}{ \sin 7x} \cdot \frac{\cos 7x}{1}\right]$$

$$=\lim_{x \longrightarrow 0}\left[\frac{\frac{\sin 4 x}{1}}{\frac{\sin 7x}{1}} \cdot \cos 7x\right] = \lim_{x \longrightarrow 0}\left[\frac{\frac{4x}{4x}\cdot\frac{\sin 4 x}{1}}{\frac{7x}{7x}\cdot\frac{\sin 7x}{1}} \cdot \cos 7x\right]$$

$$=\lim_{x \longrightarrow 0}\left[\frac{4x\cdot\frac{\sin 4 x}{4x}}{7x\cdot\frac{\sin 7x}{7x}} \cdot \cos 7x\right] = \lim_{x \longrightarrow 0}\left[\frac{4x}{7x}\cdot\frac{\frac{\sin 4 x}{4x}}{\frac{\sin 7x}{7x}} \cdot \cos 7x\right]$$

$$= \lim_{x \longrightarrow 0}\left[\frac{4}{7}\cdot\frac{\frac{\sin 4 x}{4x}}{\frac{\sin 7x}{7x}} \cdot \cos 7x\right] = \frac{4}{7}\cdot\lim_{x \longrightarrow 0}\left[\frac{\frac{\sin 4 x}{4x}}{\frac{\sin 7x}{7x}} \cdot \cos 7x\right]$$

$$= \frac{4}{7}\cdot\frac{\displaystyle\lim_{x \longrightarrow 0}\frac{\sin 4 x}{4x}}{\displaystyle\lim_{x \longrightarrow 0}\frac{\sin 7x}{7x}} \cdot \lim_{x \longrightarrow 0}\cos 7x = \frac{4}{7}\cdot \frac{1}{1}\cdot 1 = \frac{4}{7}\,.$$

Boy, that was quite a bit of work! Let's just use the magic "fact" to save time. The point of writing that in excruiciating detail is to show what is happening behind the scenes. It is very basic and fundamental but certainly not magic.

I know "crossing out" $\sin x$ seems easier but you must swear off of it now for the sake of all that is sacred.

Hut
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