Question

Please help will fan and medal, find the limit using l'hospital's rule (xe^3x-x)/(1-cos2x)

Answer 1

x approaches 0?

Answer 2

my bad im sorry the limit does approach 0

Answer 3

\[\lim_{x \rightarrow 0} \frac{xe^{3x}-x}{1-\cos(2x)}\]

Answer 4

yes

Answer 5

so we do have 0/0 when x approaches 0
so that does mean we can apply l'hospital

Answer 6

yes

Answer 7

have you tried differentiating top and bottom w.r.t. x ?

Answer 8

I can do the bottom

Answer 9

and I know -x=-1 for the top

Answer 10

\[(xe^{3x}-x)' \\ =(xe^{3x})'-(x)' \text{ by difference rule } \\ =(x)'e^{3x}+x(e^{3x})'-(x)' \text{ by product rule } \\ =...\]

Answer 11

so you are left differentiating x and e^(3x)

Answer 12

yes so it would be? e^3x+xe^3x-1?

Answer 13

almost you have differentiated e^(3x) just a bit incorrectly

Answer 14

Oh wait would it be 3e^3x?

Answer 15

yep

Answer 16

\[\lim_{x \rightarrow 0}\frac{e^{3x}+3xe^{3x}-1}{2 \sin(2x)}\]

Answer 17

okay I'm good to this step

Answer 18

nothing is obvious to me here
but when I do plug in 0 again I do get 0/0
so I would try applying l'hospital again

Answer 19

omg!!! this is going to be really messy

Answer 20

but...

Answer 21

notice the bottom will be 4 cos(2x)
and 4 cos(2x) evaluated at x=0 is not 0
:)
so we are almost done

Answer 22

yay :)

Answer 23

try diff the numerator

Answer 24

okay

Answer 25

well I know the first part is 3e^3x

Answer 26

for the second term you will have to apply product rule

Answer 27

Okay

Answer 28

\[(3xe^{3x})' \\ (3x)' e^{3x}+3x(e^{3x})' \text{ by product rule }\]

Answer 29

so? 3(x)'(e^3x)+(e^3x)'(x)

Answer 30

almost you missed the 3 in the second term you just wrote

Answer 31

like a factor of 3 that is

Answer 32

oops

Answer 33

so than you will get 3e^3x+9xe^3x?

Answer 34

one sec let me check

Answer 35

\[(e^{3x}+3x e^{3x}-1)' \\ =(e^{3x})'+(3xe^{3x})'-(1)' \\ =3e^{3x}+3e^{3x}+3x \cdot 3e^{3x}-0 \\ =6e^{3x}+9xe^{3x}\]

Answer 36

oh you mean just for the second term you are right

Answer 37

and then we add the derivative from that first term and we are good to go

Answer 38

\[\lim_{x \rightarrow 0}\frac{6e^{3x}+9xe^{3x}}{4 \cos(2x)}\]

Answer 39

now you are sorta done with the calculus part

Answer 40

just need to plug in 0

Answer 41

okay

Answer 42

e^0 is 1 right?

Answer 43

yep

Answer 44

I got 3/2

Answer 45

:)

Answer 46

looks great
you get 6/4 which reduces to 3/2
great job

Answer 47

thank you :) can I ask you about one more but I'll put it in a new tab, it was one we had to do in class but I didn't get it right

Answer 48

sure

Answer 49

Thank you :)