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

**PhizKid** · September 18th 2012, 03:14 AM

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.

**Prove It** · September 18th 2012, 03:32 AM

Originally Posted by PhizKid

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.

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

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

**PhizKid** · September 18th 2012, 03:37 AM

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).

**PhizKid** · September 18th 2012, 03:54 AM

I used substitution for tan(x) = u and got tanx/sinx = 1/cosx = 1/1

**Prove It** · September 18th 2012, 04:02 AM

Originally Posted by PhizKid

I used substitution for tan(x) = u and got tanx/sinx = 1/cosx = 1/1

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

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

So in your case

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

**emakarov** · September 18th 2012, 04:04 AM

I already typed a similar solution.

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

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

**SworD** · September 18th 2012, 07:17 AM

With intuition you can get the answer from just a glance. Tan(x) behaves like x for values near 0, so the limit is equivalent to sin(x)/sin(x).

**topsquark** · September 18th 2012, 04:10 PM

Originally Posted by SworD

With intuition you can get the answer from just a glance. Tan(x) behaves like x for values near 0, so the limit is equivalent to sin(x)/sin(x).

This is a tempting method, but it sort of begs the question: How do you know that $\lim_{x \to 0}{tan(x)} \approx x$ ? 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

**SworD** · September 18th 2012, 04:35 PM

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

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

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