I got a doubt in limits see below for my question

- anonymous

I got a doubt in limits see below for my question

- chestercat

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- anonymous

\[\lim_{x \rightarrow 0}\]\[sinax / tanbx\]

- anonymous

its 0/0 form

- anonymous

This is the actual question \[\lim_{x \rightarrow 0} sinax/tanbx\]

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## More answers

- anonymous

differentiate top and bottom acos(ax) / b sec^2 (bx)

- anonymous

sub x=0

- anonymous

answer = (a/b)

- anonymous

multiply and divide by bx/ax

- anonymous

I don't know differntiation and u mean that i should multiply the numerator by ax/ax and denominator by bx/bx????

- anonymous

I dont even know what him was on bout either lol

- anonymous

if you havent done derivatives then your teacher shouldnt be giving you this question !

- anonymous

this can be done without derivatives

- anonymous

that's wat i am saying him1618 you are right

- anonymous

how? , and its much easier with derivatives, 2lines of work

- anonymous

its not pinching theorm

- anonymous

yeah its easier..no doubt 110 times out of 100 id use derivatives, but the fact is that if hes doing limits fundamentally its imperative for him to know the fundamental steps..rather than shortcuts

- anonymous

so i guess he wants a legitimate fundamental method which uses standard limit definitions

- anonymous

exactly now can u pls explain

- anonymous

first divide above and below by x

- anonymous

then make a form like this
\[[\sin( ax )/ax] /[\tan (bx)/bx] \] x a/b

- anonymous

the book i am goin through says multiply the numerator by ax/ax and the denominator by bx/bx

- anonymous

yeah thats right..dont look at the solns..which book is this?

- anonymous

u understood wt i did?

- anonymous

nope the book is my coaching book

- anonymous

i did not understand wat wud happen if i divided by x????

- anonymous

its the same thing

- anonymous

ok divide sin ax by ax and tan bx by bx

- anonymous

and then you will have to multiply by a/b also, to keep it the same..understand?

- anonymous

so sinax by ax, with x tending to zero is 1, and so is tanbx by bx
so the ans is 1 x a/b
a/b

- anonymous

samjha?

- anonymous

yeh I remember that way , just the way the question looked put me off lol

- anonymous

also, I just differentiate top and bottom etc, its so easy and there no brain required

- anonymous

yeah everyone does that..me too.bt the kid wanted a method...i know no one in the world would not use derivatives for this..even the kid will start using em after a couple of months

- anonymous

ok ab question yeh hai ki jo ax aur bx hum multiply aur divide karte hain woh idhar kya kar rahen hai???

- anonymous

tujhe pata hai sinx / x is 1 when x tends to zero

- anonymous

how do i k now dat is it a rule or smthing??

- anonymous

its the first rule you learn in limits

- anonymous

let me check

- anonymous

sinx / x =1 and tanx / x is 1 when x tends to zero

- anonymous

k yes got it!!!!!!

- anonymous

so now if you have sin kx /x with x tending to zero and k any integer, whts the limit

- anonymous

is it ax/bx??

- anonymous

sin kx / x with x tending to zero??answer kya hai?

- anonymous

1

- anonymous

no its k

- anonymous

multiply and divide by k

- anonymous

yeah its k sorry yaar!!!!!!

- anonymous

so in the same way u treat ur question as two different limits and then take their ratio...

- anonymous

but if we do dis we get \[\sin ax/ax * ax = 1*ax = ax\]

- anonymous

but then u have bx below..cancel x..u get a/b

- anonymous

can i cancel (sin ax)/ax ka x???? to get a i don't think so!!!!

- anonymous

vaise u cant..but when uve got b in the denominator u can

- anonymous

how and y can i do dat??? any spl reason behind it???

- anonymous

abe...............sinax / ax / tanbx / bx x a/b

- anonymous

is it the same as sinax/tanbx

- anonymous

yups

- anonymous

to evaluate sinax/tanbx u need to form sinax/ ax/tanbx/bx....to make the expression same uve got to multiply by a/b

- anonymous

luks like i got it the sin x / x = 1 wala way!!!!!!

- anonymous

i got ax / bx where i can cancel x

- anonymous

is this tareeka ok!!!!!!!!!

- anonymous

sin ax / tan bx = sin(ax) x ax
---------
ax
----------
tan (bx) x bx
---------
bx

- anonymous

theek hai?

- anonymous

is the ax between the 2 xes for multiplying

- anonymous

abe that x is multiply

- anonymous

if so tab theek hai got it!!!!!!

- anonymous

thanx got it!!!!!!!!

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