# Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 17|

#### $\lim\limits_{x\to 0}\left(\frac{sin\left(x\right)sin\left(2x\right)}{1-cos\left(x\right)}\right)$

SOLUTION:

Direct substitution of $x=0$ gives the indeterminate form $\frac{0}{0}$. Therefore, we should apply trigonometric identities.

We know that $sin\left(2x\right)=2sin\left(x\right)cos\left(x\right)$, so we can rewrite the original function as

$\lim\limits_{x\to 0}\left(\frac{sin\left(x\right)sin\left(2x\right)}{1-cos\left(x\right)}\right)=\lim\limits_{x\to 0}\left(\frac{sin\left(x\right)2\left(sin\left(x\right)cos\left(x\right)\right)}{1-cos\left(x\right)}\right)$

$=2\cdot \lim\limits_{x\to 0}\left(\frac{sin^2\left(x\right)cos\left(x\right)}{1-cos\left(x\right)}\right)$

We also know the pythagorean identity, $sin^2\left(x\right)=1-cos^2\left(x\right)$. So,

$=2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1-cos^2\left(x\right)\right)cos\left(x\right)}{1-cos\left(x\right)}\right)$

$=2\cdot \lim\limits_{x\to 0}\left(\frac{\left(1+cos\left(x\right)\right)\left(1-cos\left(x\right)\right)cos\left(x\right)}{1-cos\left(x\right)}\right)$

$=2\cdot \lim\limits_{x\to 0}\left(\left(1+cos\left(x\right)\right)cos\left(x\right)\right)$

$=2\cdot \left(\left(1+cos\left(0\right)\right)cos\left(0\right)\right)$

$=2\cdot \left(\left(1+1\right)\cdot 1\right)$

$=4$