Limit Comprehension Question.

**KhanDisciple** · January 13th 2013, 05:52 PM

The limit problem is as follows: $\lim_{x\to\0}\frac{11x}{tan21x}$ .
Some thoughts I had about solving this were to express the inverse of tangent as a fraction of cosine divided by sine: $\frac{cos11x}{sin21x}$ . By doing this I thought that I could then use the $\lim_{x\to\0}\frac{sin(x)}{x}=1$ theorem(?) to multiply the numerator by 21 yielding $\frac{21x}{sin21x}$ which equals one. But I don't really know what to do with cosine and this is where I am stuck. The answer in the book is $\frac{11}{21}$ . If someone could help point me in the right direction I would really appreciate it.

(P.S., sorry if this is in the wrong forum)

**Barioth** · January 13th 2013, 06:35 PM

You solved it already.

since $tan(21x) = \frac{\sin(21x)}{\cos(21x)}$ you end up with

$\lim_{x\to\0}{\frac{11x*cos21x}{sin21x}}$
$= 11* \lim_{x\to\0}{\frac{ 21x *\cos(21x)}{21*\sin(21x)}}$
$= 11* \lim_{x\to\0}{\frac{1* \cos(21x)}{21}}$ .. .. .. .. Since you said $\lim_{x\to\0} {\frac{21x}{sin(21x} = 1$
$= 11*\frac{1}{21} * \lim_{x\to\0}{\cos(21x)}$ .. .. .. .. since $\cos(21*0) = 1$
$=\frac{11}{21}$

Have a good day

**Soroban** · January 13th 2013, 06:49 PM

Hello, KhanDisciple!

You have the right idea.
We just need some Olympic-level gymnastics.

$\lim_{x\to0}\frac{11x}{\tan(21x)}$

We have: . $\frac{11x}{\tan(21x)} \;=\; \frac{11x}{\frac{\sin(21x)}{\cos(21x)}} \;=\;\frac{11x\cos(21x)}{\sin(21x)}$

Multiply by $\frac{21}{21}:\;\;\frac{21}{21}\cdot\frac{11x\cos( 21x)}{\sin(21x)} \;=\;\frac{11}{21}\cdot\frac{21x}{\sin(21x)}\cdot \cos(21x)$

Therefore: . $\lim_{x\to0}\left[\frac{11}{21}\cdot\frac{21x}{\sin(21x)}\cdot\cos(2 1x)\right] \;=\;\frac{11}{21}\cdot 1 \cdot 1 \;=\;\frac{11}{21}$

**Prove It** · January 13th 2013, 08:23 PM

Originally Posted by KhanDisciple

The limit problem is as follows: $\lim_{x\to\0}\frac{11x}{tan21x}$ .
Some thoughts I had about solving this were to express the inverse of tangent as a fraction of cosine divided by sine: $\frac{cos11x}{sin21x}$ . By doing this I thought that I could then use the $\lim_{x\to\0}\frac{sin(x)}{x}=1$ theorem(?) to multiply the numerator by 21 yielding $\frac{21x}{sin21x}$ which equals one. But I don't really know what to do with cosine and this is where I am stuck. The answer in the book is $\frac{11}{21}$ . If someone could help point me in the right direction I would really appreciate it.

(P.S., sorry if this is in the wrong forum)

If you didn't want to perform the Olympic Level gymnastics, L'Hospital's Rule simplifies things greatly since this goes to $\displaystyle \begin{align*} \frac{0}{0} \end{align*}$ .

$\displaystyle \begin{align*} \lim_{x \to 0}\frac{11x}{\tan{\left( 21x \right)}} &= \lim_{x \to 0}\frac{\frac{d}{dx} \left( 11x \right)}{\frac{d}{dx} \left[ \tan{\left( 21x \right)} \right]} \\ &= \lim_{x \to 0} \frac{11}{21\sec^2{\left( 21x \right)}} \\ &= \frac{11}{21} \end{align*}$

**KhanDisciple** · January 13th 2013, 10:28 PM

Thanks a lot guys, you are the best!

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