Question

Calculus,

Can you please help me solve (a) and (b) ?
https://www.dropbox.com/s/xslzq5tpy26zpkd/Screenshot%202013-11-02%2021.04.51.jpg

Answer 1

@hartnn

Answer 2

@AravindG

Answer 3

so did you apply l.h rule ?

Answer 4

I did

Answer 5

And I got tanxcosx/sinxsec^2x

Answer 6

yes,
now simplify that

Answer 7

tan cos is just sin

Answer 8

oh so its 1

Answer 9

yep

Answer 10

how about (b) ?

Answer 11

what would your first step be ?

Answer 12

as far as I understand is something ^inf form

Answer 13

taking log on both sides

Answer 14

what the i.h rule

Answer 15

L = lim ....
log L = lim log ...

Answer 16

it's L.H. and he means L'hospital's

Answer 17

^

Answer 18

I got ln(y) = ln(1 + 2/x) / (1/x)

Answer 19

b is easy just put in infinity or a large number like 1,000,000 and see what it is close too

Answer 20

yes, you can differentiate now...0/0 gorm

Answer 21

form

Answer 22

is that 1 too ?

Answer 23

yes

Answer 24

but wait

Answer 25

I still have an ln(y) here

Answer 26

1+0 to any power is still just 1

Answer 27

you got 1 ??? or 2 ?

Answer 28

1 :(

Answer 29

you should get 2

Answer 30

oh I messed up badly hold

Answer 31

num diff = -2/x^2

Answer 32

true, its 2

Answer 33

so , ln y = 2

Answer 34

y= ... ?

Answer 35

e^2 !

Answer 36

yaaay

Answer 37

;p

Answer 38

lemme try © too just to be safe

Answer 39

sure :)

Answer 40

inf/0 is not eligible for LH right ?

Answer 41

how r u getting inf for numerator ?

Answer 42

I did it like this - 1/six(x) / x

Answer 43

nah its stupid, hold on , lol

Answer 44

let me make your life easier
1/x - cosec x
= (1-x cosec x )/x

Answer 45

oh so thats inf

Answer 46

wot ?

Answer 47

treat 0*inf as 0

Answer 48

x cosec x this is 0

Answer 49

so we have 1/x

Answer 50

aka inf ?

Answer 51

num = 0
denom = 0
0/0
apply L.H
.-.

Answer 52

man, (1-x cosec x )/x <------ xcsc(x) is 0 right ?

Answer 53

yes

Answer 54

so we are left with 1/x right ?

Answer 55

nvm, I got it

Answer 56

i was asking for the form

Answer 57

whether you can apply LH or not

Answer 58

actually I didn't got it

Answer 59

ah

Answer 60

I don't think we need to apply it ?

Answer 61

is there 0/0 form ?

Answer 62

there is (1 + 0 / 0 ) form

Answer 63

ah if I break it there is 0/0

Answer 64

that's what you mean ?

Answer 65

i am stupid , sorry
do it this way only
1/x - 1/sin x
= (sin x - x) / (x sin x)

Answer 66

oh hm

Answer 67

the result is 0 ?

Answer 68

yes!

Answer 69

:D thanks a lot

Answer 70

welcome ^_^