# Question #eefeb

Sep 28, 2015

By using trig identities and algebraic manipulation, together with L'Hospitals Rule and other limit laws, we eventually get this limit equal to $\frac{1}{2}$.

#### Explanation:

By a trig identity, $\sin 2 x = 2 \sin x \cos x$

Taking out a common denominator and simplifying algebraically, we get

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

This is an indeterminant limit form of type $\frac{0}{0}$ and therefore we may use L'Hospital's Rule to evaluate it to obtain

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

$= {\lim}_{x \to 0} \left(\frac{1 - {\sin}^{3} x - 4 \sin x {\cos}^{2} x}{2 {\cos}^{2} x - 2 \sin x \cos x}\right)$

$= \frac{1}{2}$

Sep 28, 2015

See the explanation section.

#### Explanation:

To find ${\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} \left(2 x\right) - \sin x\right)$,

we'll need the fundamental trigonometric limit: ${\lim}_{\theta \rightarrow 0} \sin \frac{\theta}{\theta} = 1$

in the reciprocal form:
${\lim}_{\theta \rightarrow 0} \frac{\theta}{\sin} \theta = 1$

${\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} \left(2 x\right) - \sin x\right) = {\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} \left(2 x\right)\right) - {\lim}_{x \rightarrow 0} \sin x$

$= {\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} \left(2 x\right)\right) - 0$

So the challenge is finding ${\lim}_{x \rightarrow 0} \frac{x}{\sin} \left(2 x\right)$

Solution 1
${\lim}_{x \rightarrow 0} \frac{x}{\sin} \left(2 x\right) = {\lim}_{x \rightarrow 0} \frac{1}{2} \frac{2 x}{\sin} \left(2 x\right)$

$= \frac{1}{2} {\lim}_{x \rightarrow 0} \frac{\left(2 x\right)}{\sin} \left(2 x\right)$

$= \frac{1}{2} \left(1\right) = \frac{1}{2}$

We used the fact that as $x \rightarrow 0$, we also have $2 x \rightarrow 0$. We could formalize the substitution $\theta = 2 x$ is we want to be precise.

(The technique in solution 1 also works for ${\lim}_{x \rightarrow 0} \frac{x}{\sin} \left(5 x\right) = \frac{1}{5}$.)

Solution 2

Use $\sin \left(2 x\right) = 2 \sin x \cos x$ to rewrite.

${\lim}_{x \rightarrow 0} \frac{x}{\sin} \left(2 x\right) = {\lim}_{x \rightarrow 0} \frac{x}{2 \sin x \cos x}$

$= {\lim}_{x \rightarrow 0} \frac{x}{2 \sin x \cos x}$

$= {\lim}_{x \rightarrow 0} \frac{1}{2} \frac{x}{\sin} x \frac{1}{\cos} x$

$= \frac{1}{2} \left[{\lim}_{x \rightarrow 0} \frac{x}{\sin} x\right] \left[{\lim}_{x \rightarrow 0} \frac{1}{\cos} x\right]$ (if both limits exist)

$= \frac{1}{2} \left[1\right] \left[\frac{1}{1}\right] = \frac{1}{2}$

(Unless you know the formula for $\sin \left(5 x\right)$, this solution method will not work for ${\lim}_{x \rightarrow 0} \frac{x}{\sin} \left(5 x\right)$.)