# Question #1f9ed

Dec 15, 2017

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

#### Explanation:

Given ${\lim}_{x \rightarrow 0} \frac{x \cos x + x}{\sin x \cos x}$ first factor the numerator:

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

Now rearrange a little to make things more explicit:

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

as long as both limits exist we can change this to a product of two limits:

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

Since ${\lim}_{x \rightarrow 0} \left(\sin \frac{x}{x}\right) = 1$ its reciprocal will also be 1, that is ${\lim}_{x \rightarrow 0} \left(\frac{x}{\sin} x\right) = 1$, so:

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

The remaining limit can be evaluated by direct substitution:

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

Dec 15, 2017

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

#### Explanation:

${\lim}_{x \to 0} \frac{x \cos x + x}{\sin x \cos x}$

$\equiv {\lim}_{x \to 0} \frac{x \cos x}{\sin x \cos x} + {\lim}_{x \to 0} \frac{x}{\sin x \cos x}$

$\equiv {\lim}_{x \to 0} \frac{x}{\sin x} + {\lim}_{x \to 0} \frac{x}{\sin x \cos x}$

$\equiv {\lim}_{x \to 0} \frac{x}{\sin x} + {\lim}_{x \to 0} \frac{x}{\sin x} \cdot {\lim}_{x \to 0} \frac{1}{\cos} x$

Since ${\lim}_{x \to 0} \frac{\sin x}{x} = 1$ then ${\lim}_{x \rightarrow 0} \frac{x}{\sin} x = 1$

$\equiv 1 + 1 \cdot {\lim}_{x \to 0} \frac{1}{\cos} x$

$\equiv 1 + {\lim}_{x \to 0} \frac{1}{\cos} x$

${\lim}_{x \to 0} \frac{1}{\cos} x = 1$

$\equiv 1 + 1 = 2$

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