Question

how to find the limit of csc 3x/ cots as x approaches 0

Answer 1

\[\lim_{x \rightarrow 0} \frac{ \csc 3x }{ \cot } \]
is that what you mean?

Answer 2

what is "cots" in the bottom?

Answer 3

I think it is meant to be x, not s. (These letters are close enough to each other on the keyboard to make that mistake. Or at least I make these type of mistakes fairly often.)
\[\Large \color{blue}{\lim_{x \rightarrow 0}~~\frac{\csc (3x)}{\cot(x)}}\]
\[\Large \color{blue}{\lim_{x \rightarrow 0}~~\csc (3x)\tan(x)}\]
\[\Large \color{blue}{\lim_{x \rightarrow 0}~~\csc (3x)\times \lim_{x \rightarrow 0}~\tan(x)}\]
\[\Large \color{blue}{\lim_{x \rightarrow 0}~~\csc (3x)\times 0 }\]
\[\Large \color{blue}{undefined \times 0 }\]

Answer 4

it is undefined.

Answer 5

You can re-write csc(3x) ion different forms, but you will still get an undefined value for the limit, something/0.

Answer 6

we can use L'Hopital's Rule for indeterminate forms such as 0/0.
\[ \lim_{x\rightarrow 0} \frac{\tan x}{\sin 3x} = \lim_{x\rightarrow 0} \frac{\frac{d}{dx}\tan x}{\frac{d}{dx} \sin 3x} \]

Answer 7

If you do not know calculus or L'Hopital's Rule, you can use trig identities.
See https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae
(or expand sin(3x) as sin(x + 2x), and then expand the 2x terms). either way, you can write
\[ \lim_{x\rightarrow 0} \frac{\tan x}{\sin 3x} = \lim_{x\rightarrow 0} \frac{\sin x}{\cos x(-4 \sin^3 x+3 \sin x)} \\=\lim_{x\rightarrow 0} \frac{1}{\cos x(-4 \sin^2 x+3 )}
\]