Question

Help please.Evaluate.
lim x-0 {(sin2x +3x)/( 2x +sin3x)]
Please do it with steps

Answer 1

You can use identities

Answer 2

is that sin(2x) or sin^(2)(x)

Answer 3

sin2x

Answer 4

Im confused, please make me see your steps

Answer 5

yeah I guess you could use L'H rule

Answer 6

Let me reformulate:

Answer 7

Take the derivative of the top and the bottom. Treat the neumorator and denominator as two seperate non fractional equations

Answer 8

to use L'H rule

Answer 9

lim x-0 {(sin2x +3x)/( 2x +sin3x)] = L'H
=
lim (2cos(x) + 3)/(2 + 3cos(x))
x->0

Answer 10

\[\lim_{x \rightarrow 0}[(\sin{2x} + 3x)/(2x + \sin{3x})]\]By L.H., we get:\[\lim_{x \rightarrow 0}[(2\cos{2x} + 3)/(3\cos{3x} + 2)]\]Try it out now

Answer 11

= 5/5 = 1

Answer 12

If he isn't allowed to use L'H rule perhaps he is better of using trig Identities

Answer 13

Indeed, but they are a pain to use on an exam, except for the identity one with sin^2 + cos^2, :-)

Answer 14

sorry I mean
lim ln(x)ln(1/x)
x->0+

Answer 15

I have done this ..PLEASE PLEASE PLEASE VERIFY MY STEPS, TELL ME IF IM RIGHT OR WRONG

Answer 16

If you want to use trig identites rememebr that

sin(2x) = 2sin(x)cos(x)

I can't recall what the sin(x/y) -

Answer 17

no right sorry :)

Answer 18

7/2 / 7/2 = 1

Answer 19

M i WRONG?

Answer 20

No you got 1 so you are right I have no idea what you did but you got the right answer

Answer 21

thats wht I was telling PLEASE CHECK MY STEPS

Answer 22

You need to use Identites sin(2x) = 2sin(x)cos(x)

then there is another identity that is something like sin(xy) = sin(x)cos(y) - sin(x)cos(y)

Answer 23

These are the identites you need to use
sin(A+B) = sinAcosB + cosAsinB
sin(2x) = 2sin(x)cos(x)

To solve this

sin(3x) = sin(2x + x)

Answer 24

or we can do like this
lim x-0 {(sin2x)/x +3}/ {2 +(sin3x)/x}
put sin2x/x = (2sin2x)/2x = 2and (3sin3x)/3x =3bcz
lim x-0 (sinx)/x

so we will get (2+3)/3+2) = 1

Answer 25

@rahul33333 Almost right, but d(sin3x)/dx = 3sin(3x) not sin(3x)/3. Other than that, it's correct what you did.

Answer 26

I mean the same for cos(2x)

Answer 27

ohhhhooo, I did a mistake yes yes. thanks , MEANS I LEARN L HOSPITAL RULE !!! wow

Answer 28

Kudos :-)

Answer 29

Crud you need also this identity cos(2x) = cos²(x) - sin²(x)

we can get rid of those ugly trig functions by

(2cos(x)sin(x) +3x)/(2x + sin(2x)cos(x) +sin(x)cos(2x)))
=
(2cos(x)sin(x) + 3x)/(2x + 2sin(x)cos(x)cos(x) + sin(x)(cos^(2)(x) - sin^(2)(x)))

Answer 30

fromer was a better way, anyway thanks

Answer 31

lol have fun simplifying that though :)