Question

lim x->0 of tan6x/sin2x ?

Answer 1

3

Answer 2

It would be helpful if you showed your work! thankss

Answer 3

I know that tan 6x turns into sin6x/cos6x. and then I'm stuck

Answer 4

Do you know L'Hopital's Rule?

Answer 5

I heard of it but forgot what it actually is.

Answer 6

Technically, kind of. It is a method you can use when both numerator and denominator are zero. All you have to do is differentiate both numerator and denominator.

Answer 7

= lim x->0 dtan(6x)/dx/dsin(2x)/dx : = lim x->0 3/(cos(2x)cos6x)^2)

factor out constants
3 (lim x->0 1/(cos(2x)cos(6x)^2)

the limit of a quotient is the quotient of the limits:
the limit of a constant is the constant:
3/(lim x->0 cos(2x)cos(6x)^2)

the limit of a product is the product of the limits:

3/ (lim x->0 cos (2x))(lim x->0 cos(6x)^2)

Answer 8

Using the continuity of cos(x) at x=0 write lim x->0 cos(2x) as cos(lim x->0 2x)

Answer 9

Let me show you.

Since \[\Large \lim_{x \rightarrow 0}\frac{\tan 6x}{\sin 2x} = \frac{0}{0}\]Applying L'Hopital's Rule: \[\Large \lim_{x \rightarrow 0}\frac{\tan 6x}{\sin 2x} = \lim_{x \rightarrow 0}\frac{\frac{d}{dx}\tan 6x}{\frac{d}{dx}\sin 2x}\]

Answer 10

And \[\large \frac{d}{dx} \tan 6x = 6 \sec^2 6x\]\[\large \frac{d}{dx} \sin 2x = 2 \cos 2x\]

Answer 11

OMG there is SO much to type!

3/cos(2(lim x->0 x)) cos (lim x->0 6x)^2

Answer 12

so I take the derivative of the numerator and denominator separately

Answer 13

chain rule?

Answer 14

Now we have \[\large \lim_{x \rightarrow 0} \frac{ 6 \sec^2 6x }{ 2 \cos 2x } = \lim_{x \rightarrow 0} \frac{ 6 }{2 \cos 6x \cos 6x \cos 2x }\]

Answer 15

"so I take the derivative of the numerator and denominator separately"

Yes, that's correct.

"chain rule?"

Yes.

Answer 16

Thank you guys so much! Got it :)

Answer 17

Glad we helped.

Answer 18

One second. I just made something that will make everything very clear.

Answer 19

No need to reply, my friend. tomefrome got it.

Answer 20

Thanks, though.

Answer 21

Thanks! I see it's a different approach but I'll take a look :D