Originally Posted by

**KhanDisciple** The limit problem is as follows: $\displaystyle \lim_{x\to\0}\frac{11x}{tan21x}$.

Some thoughts I had about solving this were to express the inverse of tangent as a fraction of cosine divided by sine: $\displaystyle \frac{cos11x}{sin21x}$. By doing this I thought that I could then use the $\displaystyle \lim_{x\to\0}\frac{sin(x)}{x}=1$ theorem(?) to multiply the numerator by 21 yielding $\displaystyle \frac{21x}{sin21x}$ 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 $\displaystyle \frac{11}{21}$. 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)