Question

Find the limit as x is approaching 0 of cot(3x)sin(6x)

Please tell me where I am going wrong in my work, which I will post below in the comments.

Answer 1

So this is the question:
\[\lim_{x \rightarrow 0} \cot(3x)\sin(6x)\]

*I made it a fraction, preparing it for L'Hoptals.
\[\sin(6x)/\tan(3x)\]

*I do L'Hopitals, taking derivative!
\[6\cos(6x)/3\csc^2(3x)\]

*simplify
\[2\cos(6x)/\csc^2(3x)\]

No I dunno what in the world to do next :(

Answer 2

What in the world...
WolframAlpha took the derivative of sin(6x)/tan(3x), and somehow got:

\[2\cos(3x)^2\cos(6x)\]

??? How wolfram alpha!

Answer 3

o
\[\lim_{x \rightarrow 0}\frac{\cos(3x)}{\sin(3x)}\sin(6x) \frac{\frac{1}{6x}}{\frac{1}{3x}} \cdot \frac{\frac{6}{1}}{\frac{3}{1}}\]
\[\lim_{x \rightarrow 0}\cos(3x) \cdot \frac{1}{\frac{\sin(3x)}{3x}} \cdot \frac{\sin(6x)}{6x} \cdot \frac{6}{3}\]
\[\cos(3 \cdot 0) \cdot \frac{1}{1} \cdot 1 \cdot \frac{6}{3}\]
\[1 \cdot 1 \cdot 1 \cdot 2=2\]

Answer 4

Okay? so i am still confused? I don't know why you did what you did. From the first step...

Answer 5

so you don't know this:
\[\lim_{u \rightarrow 0}\frac{\sin(u)}{u}=1 \]
?

Answer 6

No. What is that? i've never seen that beforee!

Answer 7

thats weird
they teach that before l'hospital's rule

Answer 8

like way before
we can prove it by squeeze thm

Answer 9

Yeah i know squeeze theroem. But i need to know how to do it the normal, intedned way. L'hoptials.

Answer 10

why is that the normal way?

Answer 11

the way i showed you is easiest

Answer 12

i just multiplied by 1/1
so i could apply that lim(u->0) sin(u)/u=1

Answer 13

Wait, I also didn't remember that cos/sin = cot.

About this sinu/u thing? Are there any others?

Answer 14

I thought cot=1/tan...

Answer 15

\[\lim_{x \rightarrow 0}\frac{\cos(x)-1}{x}=0\]

Answer 16

we can prove that multiplying top and bottom by the conjugate of the top
and then applying lim(x->0) sin(x)/x=1

Answer 17

I need to do this using lhoptials.

We are learning how to use lhoptals. The whole point of what I'm doing is to learn the application of the rule.

Answer 18

\[\cot(x)=\frac{adj}{opp} ; \frac{\cos(x)}{\sin(x)}=\frac{\frac{adj}{hyp}}{\frac{opp}{hyp}}=\frac{adj}{opp}\]
\[=>\cot(x)=\frac{\cos(x)}{\sin(x)}\]

Answer 19

\[\lim_{x \rightarrow 0}\frac{\sin(6x)}{\tan(3x)}\]
so we have 0/0 we can apply l'hospital rule
\[\lim_{x \rightarrow 0}\frac{6 \cos(6x)}{3 \sec^2(3x)}\]
now just plug in 0

Answer 20

\[\frac{6(1)}{3(1^2)}=2\]

Answer 21

Okay so i get were the cos/sin comes from. SO let me try it this way now:

\[\cos(3x)/\sin(3x)*sin(6x)\]

Simplify

\[\sin(3x)\cos(3x)\]

L'hopitals

\[[\cos(3x)(3)][-\sin(3x)(3)]\]


am i on the right track?

Answer 22

why are you making it so hard
i used l'hospital's rule with just one round of it
and then i was able to use direct substitution

Answer 23

K, looking at yours, sec.

Answer 24

wow wolfram does have a freaking long answer lol

Answer 25

so unnecessary though

Answer 26

Hey, look at my first step! I did exactly what you did!
SO i've been diogn it right?

Answer 27

I just simplified the 6/3 = 2/1

Answer 28

you do know \[\frac{d}{dx}(\tan(x))=\sec^2(x) \text{ and \not } \csc^2{x} \text{ iright ?}\]

Answer 29

Wait, so I know what a cos curve looks like. So i can deduce cos(0) np.

So the numerator is 6.
But I don't know what sec^2(0) is.. I mean I know it's 1 because I saw the answer. But I don't know the sec^2 curve by heart. Is the only solution to just memorize it like I have for cos tan sin?

Answer 30

\[\sec(x)=\frac{1}{\cos(x)}\]

Answer 31

if cos(x)=1, then sec(x) is also 1

Answer 32

I see... So cot = 1/tan AND cos/sin? Didn't realize it had to..

Okay well this makes sense now, I can't thank you enough!

Answer 33

there are so many trig identities

Answer 34

Sorry to ask but can i get help with another?

Creating the question now..