# What is the limit of (4x − sin4x)/(4x − tan4x) as x approaches 0?

May 17, 2016

1

#### Explanation:

Calling $a \left(x\right) = 4 x$, $b \left(x\right) = - \sin \left(4 x\right)$ and $c \left(x\right) = \cos \left(4 x\right)$ we
have ${\lim}_{x \to \infty} \frac{4 x - \sin \left(4 x\right)}{4 x - \tan \left(4 x\right)} \equiv {\lim}_{x \to 0} \left(\frac{c \left(x\right) \left(a \left(x\right) + b \left(x\right)\right)}{c \left(x\right) a \left(x\right) + b \left(x\right)}\right)$ but
${\lim}_{x \to {x}_{0}} \left(\frac{c \left(x\right) \left(a \left(x\right) + b \left(x\right)\right)}{c \left(x\right) a \left(x\right) + b \left(x\right)}\right) = \frac{c \left({x}_{0}\right) \left(a \left({x}_{0}\right) + b \left({x}_{0}\right)\right)}{c \left({x}_{0}\right) a \left({x}_{0}\right) + b \left({x}_{0}\right)}$ But in this case ${x}_{0} = 0$ and $c \left(x\right) = \cos \left(4 x\right)$ so $c \left({x}_{0}\right) = 1$. Putting all together we have
${\lim}_{x \to {x}_{0}} \left(\frac{c \left(x\right) \left(a \left(x\right) + b \left(x\right)\right)}{c \left(x\right) a \left(x\right) + b \left(x\right)}\right) = \frac{a \left({x}_{0}\right) + b \left({x}_{0}\right)}{a \left({x}_{0}\right) + b \left({x}_{0}\right)} = 1$

May 17, 2016

${\lim}_{x \rightarrow 0} \frac{4 x - 4 \sin x}{4 x - \tan 4 x} = - \frac{1}{2}$

#### Explanation:

${\lim}_{x \rightarrow 0} \frac{4 x - \sin 4 x}{4 x - \tan 4 x}$ has indeterminate initial form $\frac{0}{0}$.

I'll use l'Hospital's rule.

${\lim}_{x \rightarrow 0} \frac{4 x - \sin 4 x}{4 x - \tan 4 x} = {\lim}_{x \rightarrow 0} \frac{4 - 4 \cos 4 x}{4 - 4 {\sec}^{2} 4 x}$

$= {\lim}_{x \rightarrow 0} \frac{1 - \cos 4 x}{1 - {\sec}^{2} 4 x}$

The form is still $\frac{0}{0}$, so we'll use L'Hospital again. Note that to differentiate ${\sec}^{2} 4 x$ we'll use the chain rule twice.

$= {\lim}_{x \rightarrow 0} \frac{4 \sin 4 x}{- 2 \sec \left(4 x\right) \left[\sec \left(4 x\right) \tan \left(4 x\right) \left(4\right)\right]}$

$= {\lim}_{x \rightarrow 0} \frac{4 \sin 4 x}{- 8 {\sec}^{2} \left(4 x\right) \tan \left(4 x\right)}$

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

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

$= {\lim}_{x \rightarrow 0} \frac{- {\cos}^{3} \left(4 x\right)}{2}$ (if you prefer)

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

Here is the graph:

graph{(4x-sin(4x))/(4x-tan(4x) [-4.375, 3.42, -1.795, 2.1]}