Question

Evaluate of the limit.
\[\lim_{x \rightarrow 0}\frac{ \sin3x }{ 5x }\]

Answer 1

have you tried lhopitals rule

Answer 2

I didn't learn it yet..

Answer 3

hmm

Answer 4

trying asking other people while i remember this stuff

Answer 5

I think it's 5/3

Answer 6

it is not

Answer 7

@Nnesha

Answer 8

Oh ok sorry-

Answer 9

It's 3/5. I read it wrong.

Answer 10

Sorry I was wrong :/

Answer 11

I'm not sure how to get 3/5?

Answer 12

have you learned the maclaurin series

Answer 13

No, sorry..

Answer 14

The question is from limit of Trignometric Functions.

Answer 15

yeah i don't think ill be able to help because i don't remember any of this

Answer 16

@mathmate

Answer 17

you may use below limit
\[\large \lim\limits_{t\to 0}~\frac{\sin (t)}{t}=1\]

Answer 18

\[\large \lim\limits_{x\to 0}~\frac{\sin (3x)}{5x}\]

let \(t=3x\implies x = \frac{t}{3}\)
as \(x\to 0\), we have \(t\to 3*0 =0\)

Answer 19

the limit becomes :
\[\lim\limits_{t\to 0}~\frac{\sin (t)}{5(\frac{t}{3})} = \lim\limits_{t\to 0}~\frac{3\sin (t)}{5t} =\frac{3}{5} \lim\limits_{t\to 0}~\frac{\sin (t)}{t}=\frac{3}{5}\cdot 1=\frac{3}{5}\]

Answer 20

@esam2
Have you learned the squeeze theorem?

Answer 21

Yes

Answer 22

Try example 2 of
http://www.analyzemath.com/calculus/limits/squeezing.html

Answer 23

I think I got it now.

Answer 24

Great! :)