# How do you evaluate (sin(2h)(1 - cos h)) / h as h approaches 0?

Jul 19, 2016

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

#### Explanation:

Because direct substitution of $h = 0$ into this expression yields and indeterminate form of $\frac{0}{0}$, we can evaluate this limit using L'Hospital's rule.

In the case of the indeterminate form $\frac{0}{0}$, we take the derivative of the numerator and denominator.

Thus, we can write

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

$= {\lim}_{h \to 0} \frac{\frac{d}{\mathrm{dh}} \left(\sin \left(2 h\right)\right) - \frac{d}{\mathrm{dh}} \left(\sin \left(2 h\right) \cos \left(h\right)\right)}{\frac{d}{\mathrm{dh}} \left(h\right)}$

$= {\lim}_{h \to 0} \frac{2 \cos \left(2 h\right) - \left(2 \cos \left(2 h\right) \cos \left(h\right) + \sin \left(h\right) \cdot - \sin \left(h\right)\right)}{1}$

= lim_(h->0) (2cos(2h)(1-cos(h))+sin(2h)sin(h) = 0

By graphing our original function, we can see that it does in fact equal to $0$.

graph{(sin(2x)(1-cos(x)))/(x) [-1.922, 1.923, -0.963, 0.96]}

Jul 20, 2016

The reqd. limit $= 0$.

#### Explanation:

We use the Identity $\sin 2 h = 2 \sinh c o h$ and the fact

${\lim}_{h r a a 0} \frac{\sinh}{h} = 1$

The reqd. limit$= {\lim}_{h \rightarrow 0} 2 \left(\frac{\sinh}{h}\right) \cosh \left(1 - \cosh\right) = 2 \cdot 1 \cdot \cos 0 \cdot \left(1 - \cos 0\right)$

$= 2 \cdot 1 \cdot 1 \cdot \left(1 - 1\right) = 0$.