# How do you find the limit of sinx^2/(1-cosx) as x->0?

Feb 8, 2017

$= 2$

#### Explanation:

The Taylor Expansions for sine and cosine are:

sin (u) = u - u^{3}/{3!}+ \O (u^5)

and

cos(u) = 1 - u^2/(2!) + \O(u^4)

Plugging these into the limit yields:

lim_(x to 0) (x^2 - x^{6}/{3!} + \O (x^10))/(1-(1 - x^2/(2!) + \O(x^4)))

= lim_(x to 0) (x^2 - x^{6}/{3!} + \O (x^10))/( x^2/(2!) + \O(x^4))

Divide top and bottom by ${x}^{2}$:

= lim_(x to 0) (1 + \O (x^4))/( 1/(2!) + \O(x^2)) = 2

Feb 8, 2017

$\sin {x}^{2} / \left(1 - \cos x\right) = \sin {x}^{2} / \left(1 - \cos x\right) \frac{1 + \cos x}{1 + \cos x}$

$= \frac{\sin {x}^{2} \left(1 + \cos x\right)}{1 - {\cos}^{2} x}$

$= \sin {x}^{2} \cdot \frac{1}{\sin} ^ 2 x \left(1 + \cos x\right)$

$= \sin {x}^{2} / {x}^{2} \cdot {x}^{2} / {\sin}^{2} x \cdot \left(1 + \cos x\right)$

Using ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta}$, and ${\lim}_{x \rightarrow 0} \cos x - 1$ we get

${\lim}_{x \rightarrow 0} \sin {x}^{2} / {x}^{2} \cdot {x}^{2} / {\sin}^{2} x \cdot \left(1 + \cos x\right) = \left(1\right) \left(1\right) \left(1 + 1\right) = 2$