# Question 7a84b

Mar 27, 2017

${\lim}_{x \to 0} \left(\frac{1}{\sin} x - \frac{1}{\tan} x\right) = 0$

#### Explanation:

We have to evaluate the limit:

${\lim}_{x \to 0} \left(\frac{1}{\sin} x - \frac{1}{\tan} x\right)$

use the trigonometric identity: $\tan x = \sin \frac{x}{\cos} x$

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

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

Now use the formulas for the double angle, considering $x = 2 \cdot \frac{x}{2}$:

(1-cosx) = 1 - (cos^2(x/2) - sin^2(x/2)) = 1 - cos^2(x/2) + sin^2(x/2)) =2sin^2(x/2)#

$\sin x = 2 \sin \left(\frac{x}{2}\right) \cos \left(\frac{x}{2}\right)$

so that:

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

Mar 27, 2017

${\lim}_{x \rightarrow 0} \left(\frac{1}{\sin} x - \frac{1}{\tan} x\right) = {\lim}_{x \rightarrow 0} \sin \frac{x}{1 + \cos x} = 0$

#### Explanation:

$\frac{1}{\sin} x - \frac{1}{\tan} x = \frac{1}{\sin} x - \cot x = \frac{1}{\sin} x - \cos \frac{x}{\sin} x = \frac{1 - \cos x}{\sin} x$

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

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

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

$= \sin \frac{x}{1 + \cos x}$

So,

${\lim}_{x \rightarrow 0} \left(\frac{1}{\sin} x - \frac{1}{\tan} x\right) = {\lim}_{x \rightarrow 0} \sin \frac{x}{1 + \cos x}$

$= \frac{0}{1 + 1} = 0$