# How do you find the limit of (cos x - 1) / sin x^2 as x approaches 0?

Feb 19, 2016

Use the fundamental trigonometric limits and algebra.

#### Explanation:

The fundamental trigonometric limits are:

${\lim}_{x \rightarrow 0} \frac{\cos x - 1}{x} = 0$ and ${\lim}_{x \rightarrow 0} \sin \frac{x}{x} = 1$

If the problem is
$\frac{\cos x - 1}{\sin} \left({x}^{2}\right)$, then use

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

So the limit we seek is equal to ${\lim}_{x \rightarrow 0} \frac{\cos x - 1}{x} ^ 2$

Looking for a trick that might help, we it may eventually occur to us to try mulltiplying by $\frac{\cos x + 1}{\cos x + 1}$ to replace the subtraction that goes to $0$ with an addition that goes to $2$.

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

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

The limit as $x \rightarrow 0$ is $- \left(1\right) \cdot \left(1\right) \cdot \frac{1}{1 + 1} = - \frac{1}{2}$

Bonus

If the problem is intended to be $\frac{\cos x - 1}{\sin x} ^ 2$, then use the same trick to get:

$\frac{\cos x - 1}{\sin x} ^ 2 = \frac{\cos x - 1}{\sin x} ^ 2 \frac{\cos x + 1}{\cos x + 1}$

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

and, again, the limit is $- \frac{1}{2}$ (Without using fundamental trigonometric limits).

Feb 19, 2016

Alternatively, you may also use L'Hospital's Rule to still get the same answer to the limit as $- \frac{1}{2}$.

#### Explanation:

Since the limit is an indeterminant type of form $\frac{0}{0}$, we may apply L'Hospital and differentiate the top and bottom twice to eventually obtain :

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

$= {\lim}_{x \to 0} \frac{- \cos x}{- 4 {x}^{2} \sin x + 2 \cos {x}^{2}}$

$= - \frac{1}{2}$