The limit as x -> 0 of sin(tan(x)) / sin(x)

Is this supposed to be an intuitive question? Because I see no way I can manipulate this algebraically to get it into a useful form to cancel out anything. Otherwise, I don't see what the limit could be.

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- Sep 18th 2012, 03:14 AM #1

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## lim of sin(tan(x)) / sin(x)

The limit as x -> 0 of sin(tan(x)) / sin(x)

Is this supposed to be an intuitive question? Because I see no way I can manipulate this algebraically to get it into a useful form to cancel out anything. Otherwise, I don't see what the limit could be.

- Sep 18th 2012, 03:32 AM #2
## Re: lim of sin(tan(x)) / sin(x)

This goes to $\displaystyle \displaystyle \begin{align*} \frac{0}{0} \end{align*}$ so L'Hospital's Rule can be applied.

$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{\left( \tan{x} \right)}}{\sin{x}} &= \lim_{x \to 0}\frac{\frac{d}{dx}\left[ \sin{\left( \tan{x} \right)} \right]}{\frac{d}{dx}\left( \sin{x} \right)} \\ &= \lim_{x \to 0}\frac{\sec^2{x}\cos{\left(\tan{x} \right)}}{\cos{x}} \\ &= \lim_{x \to 0}\frac{\cos{\left( \tan{x}\right)}}{\cos^3{x}} \\ &= \frac{\cos{\left(\tan{0}\right)}}{\cos{0}} \\ &= \frac{1}{1} \\ &= 1 \end{align*}$

- Sep 18th 2012, 03:37 AM #3

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## Re: lim of sin(tan(x)) / sin(x)

I keep forgetting to mention that this is a Calculus 1 course, therefore we are disallowed to apply rules we haven't learned yet. I should state that explicitly in my posts from now on, or at least put it in my signature.

I can only use algebra, and some special limits we learned (particularly sinx/x and x/sinx = 1 as x-> 0).

- Sep 18th 2012, 03:54 AM #4

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- Sep 18th 2012, 04:02 AM #5
## Re: lim of sin(tan(x)) / sin(x)

I like the idea of a substitution, but you will have to rewrite $\displaystyle \displaystyle \begin{align*} \sin{x} \end{align*}$ in terms of $\displaystyle \displaystyle \begin{align*} \tan{x} \end{align*}$. From the Pythagorean Identity we have

$\displaystyle \displaystyle \begin{align*} \sin^2{x} + \cos^2{x} &\equiv 1 \\ \frac{\sin^2{x}}{\sin^2{x}} + \frac{\cos^2{x}}{\sin^2{x}} &\equiv \frac{1}{\sin^2{x}} \\ 1 + \frac{1}{\tan^2{x}} &\equiv \frac{1}{\sin^2{x}} \\ \frac{\tan^2{x} + 1}{\tan^2{x}} &\equiv \frac{1}{\sin^2{x}} \\ \frac{\tan^2{x}}{\tan^2{x} + 1} &\equiv \sin^2{x} \\ \frac{\tan{x}}{\sqrt{\tan^2{x} + 1}} &\equiv \sin{x} \end{align*}$

So in your case

$\displaystyle \displaystyle \begin{align*} \lim_{x \to 0}\frac{\sin{\left( \tan{x} \right)}}{\sin{x}} &= \lim_{x \to 0} \frac{\sin{\left( \tan{x} \right)}}{\frac{\tan{x}}{\sqrt{ \tan^2{x} + 1 }}} \\ &= \lim_{u \to 0}\frac{\sin{u}}{\frac{u}{\sqrt{u^2 + 1}}} \textrm{ after making the substitution } u = \tan{x} \\ &= \lim_{u \to 0} \frac{\sin{u}\,\sqrt{u^2 + 1}}{u} \\ &= \lim_{u \to 0}\frac{\sin{u}}{u} \cdot \lim_{u \to 0}\sqrt{u^2 + 1} \\ &= 1 \cdot 1 \\ &= 1\end{align*}$

- Sep 18th 2012, 04:04 AM #6

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## Re: lim of sin(tan(x)) / sin(x)

I already typed a similar solution.

We have $\displaystyle \cos x=\frac{1}{\sqrt{1+\tan^2x}}$ for $\displaystyle -\pi/2\le x\le\pi/2$. So

$\displaystyle \begin{align*}\lim_{x\to0}\frac{\sin(\tan x)}{\sin x}&=\lim_{x\to0}\frac{\sin(\tan x)}{\tan x\cdot\cos x}\\&=\lim_{x\to0}\frac{\sin(\tan x)}{\tan x}\sqrt{1+\tan^2x}\\ &=\lim_{t\to0}\frac{\sin t}{t}\sqrt{1+t^2}\\ &=\lim_{t\to0}\frac{\sin t}{t}\cdot\lim_{t\to0}\sqrt{1+t^2}\\ &=1\cdot1=1\end{align*}$

- Sep 18th 2012, 07:17 AM #7

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- Sep 18th 2012, 04:10 PM #8
## Re: lim of sin(tan(x)) / sin(x)

This is a tempting method, but it sort of begs the question: How do you know that $\displaystyle \lim_{x \to 0}{tan(x)} \approx x$? I am presuming that someone who is supposed to be doing the limit without an explicit mention to Calculus will not have access to the Taylor expansion of the tangent function and so cannot be used

-Dan

- Sep 18th 2012, 04:35 PM #9

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## Re: lim of sin(tan(x)) / sin(x)

You don't need taylor series. tan(x) = sin(x)/cos(x), which is close to sin(x)/1, and its been shown that sin(x) behaves like x by formally stating that

$\displaystyle \lim_{x \to 0}\frac{\sin(x)}{x} = 1$

I'm not saying everyone is expected to know this.. its just a useful strategy.

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