Question

Lim x-> 0 , tan2x/x

Answer 1

tan2x = sin2x/cos2xx

Answer 2

\[\huge\color{blue}{ \lim_{x \rightarrow 0}~~~\frac{\tan(2x)}{x} } \]

Answer 3

use L'H'S derive top and bottom,
tan(2x) ' = 2sec^2(2x)
x' = 1

Answer 4

Try not to use L'H'S because you learn that in Calculus 2. Use your trig knowledge and the theorems provided to you in Calculus 1

Answer 5

Cjr712 I have sin2x/xcos2x where do I go from there

Answer 6

Okay, @cjr712, actually your way is better.

Answer 7

@singlemom76 look at my attachment in my first post

Answer 8

I only seen a small portion of it

Answer 9

till where do you see, till " now try to do this limit" ?

Answer 10

i thought L'H'S is calculus 1.

Answer 11

Class just started 7 days ago

Answer 12

If you are at sin2x/xcos2x then multiply both the numerator and denominator by 2 so you have

lim x>0 sin2x/2x * 2/cos2x

then you can use the theorem

lim x>0 sinx/x = 1 and your left with the lim as x>o 2/cos2x

Answer 13

Can I go any further

Answer 14

@solomonzelman

Should be in the chapter that coverts Integration techniques L'Hopital's Rule, and Improper Integrals in Calculus 2.

Answer 15

the limit of x>0 sin2x/2x is 1 and then you can do the limit of 2/cos2x by pluging in a 0 for x

Answer 16

The answer is 2?

Answer 17

yes

Answer 18

I'm not allowed to give you the answer I think lol but I can confirm it :P

Answer 19

@ cjr712 want to help with another one