# Evaluate the limit. Be sure to show your reasoning. limit as x approaches 0 (sin7x / tan3x) ?

## I know that the bottom turns to sin3x/cos3x and then i multiply the recipricol in which I get the limit as x approaches 0 (sin x7)(cos3x / sin 3x) but I don't know where to go from here :(

Mar 6, 2018

$\frac{7}{3}$

#### Explanation:

$\sin \frac{7 x}{\tan} \left(3 x\right)$
=((7x)+(7x)^3/3!...)/((3x)+(1/3)(3x)^3)...(Maclaurin series)
$= \frac{7 + O \left({x}^{2}\right)}{3 + O \left({x}^{2}\right)}$ (canceling $x$'s throughout)
$\to \frac{7}{3}$ as $x \to 0$

Mar 6, 2018

$\frac{7}{3}$

#### Explanation:

use L'Hôpital's Rule (also known as hospital's rule in English)

what this rule basically means is, if we have a limit for the division of two functions which looks like $\frac{0}{0}$ or $\frac{\infty}{\infty}$ which are indeterminate forms, we can divide the derivatives of the two functions to get the answer of the limit

therefore,
${\lim}_{x \to c} f \frac{x}{g} \left(x\right) = {\lim}_{x \to c} \frac{f ' \left(x\right)}{g ' \left(x\right)}$ if f(x) and g(x) would both be 0 or infinity or negative infinity if evaluated at c

therefore, taking the derivative of both the functions, we get
${\lim}_{x \to 0} \sin \frac{7 x}{\tan} \left(3 x\right) = {\lim}_{x \to 0} \frac{\frac{d}{\mathrm{dx}} \sin \left(7 x\right)}{\frac{d}{\mathrm{dx}} \tan \left(3 x\right)}$
=

which is equal to

${\lim}_{x \to 0} \frac{7 \cos \left(7 x\right)}{3 {\sec}^{2} \left(3 x\right)}$

now since $x$ goes to 0 ,$7 x$ and $3 x$ would also approach 0

= $\frac{7 \cos \left(0\right)}{3 {\sec}^{2} \left(0\right)}$

and $\cos \left(0\right)$ and ${\sec}^{2} \left(0\right)$ are both 1

therefore, the limit is
$\frac{7}{3}$
-

Mar 6, 2018

Without using derivatives, see below.

#### Explanation:

As you have said:

$\frac{\sin 7 x}{\tan 3 x} = \frac{\sin 7 x}{\frac{\sin 3 x}{\cos 3 x}}$

$= \frac{\sin 7 x}{1} \frac{1}{\sin 3 x} \frac{\cos 3 x}{1}$

Now we'll write this so we can use
${\lim}_{t \rightarrow 0} \sin \frac{t}{t} = 1$ and ${\lim}_{t \rightarrow 0} \frac{1}{\sin} t = 1$

$= \frac{\sin 7 x}{7 x} \cdot \frac{7 x}{3 x} \frac{3 x}{\sin 3 x} \frac{\cos 3 x}{1}$

$= \frac{7}{3} \frac{\sin 7 x}{7 x} \frac{3 x}{\sin 3 x} \frac{\cos 3 x}{1}$

Now take limit as $x \rightarrow 0$ (so both $3 x$ and $7 x$ also go to $0$)

lim_(xrarr00)(sin7x)/(tan3x) = lim_(xrarr0)(7/3 (sin7x)/(7x) (3x)/(sin3x) (cos3x)/1

$= \frac{7}{3} \left(1\right) \left(1\right) \frac{1}{1} = \frac{7}{3}$