# Limit of a Function in Indeterminate Form| Differential and Integral Calculus| Feliciano and Uy| Exercise 1.3, Problem 15|

### $\displaystyle \lim_{x\to 0}\left(\frac{sin^3x}{sin\:x-tan\:x}\right)$$\displaystyle \lim_{x\to 0}\left(\frac{sin^3x}{sin\:x-tan\:x}\right)$

SOLUTION:

A straight substitution ofÂ  $\displaystyle x=\frac{\pi }{4}$ leads to the indeterminate form $\displaystyle \frac{0}{0}$ which is meaningless.

Therefore, to evaluate the limit of the given function, we proceed as follows

$\displaystyle \lim _{x\to 0}\left(\frac{sin^3x}{sin\:x-tan\:x}\right)=\lim _{x\to 0}\left(\frac{sin^3x}{sin\:x-\frac{sin\:x}{cos\:x}}\right)$

$\displaystyle =\lim _{x\to 0}\left(\frac{sin^3\left(x\right)}{\frac{sin\left(x\right)cos\left(x\right)-sin\left(x\right)}{cos\left(x\right)}}\right)$

$\displaystyle =\lim _{x\to 0}\left(\frac{sin^3\left(x\right)cos\left(x\right)}{sin\left(x\right)cos\left(x\right)-sin\left(x\right)}\right)$

$\displaystyle =\lim _{x\to \:0}\left(\frac{sin^3\left(x\right)cos\left(x\right)}{sin\left(x\right)\left(cos\left(x\right)-1\right)}\right)$

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

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

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

$\displaystyle =\lim _{x\to 0}\left(\frac{\left(1+cos\left(x\right)\right)cos\left(x\right)}{-1}\right)$

$\displaystyle =-1\cdot \lim _{x\to 0}\left(\left(1+cos\left(x\right)\right)cos\left(x\right)\right)$

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

$\displaystyle =-1\cdot \left(1+1\right)\cdot 1$

$\displaystyle =-2$