need the solution of this prob

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smileysam

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hello everyone...
I need the solution of the folowing prob.

lim (sin (x)) /(x) as x tends to 0 the answer is one ..but i don't remem as to how to prove it.

secondly as r tends to 1 suppose we have the following equation (1-r^2)/(1-r)

then we cancel out (1-r) from numerator & denom ....whats the logic as (1-r) will tend to 0 & 0 divided by 0 is undefined...

Please reply as soon as possible.
Thanks alot.
 

hi
0/0 is undefined but 1-r is not zero it only tends to zero so you can cancel it
as long as numerator and denominator are not exactly equal to zero cancellation is valid
sinx/x this you can get in any standard calculus book
bye
 

A function does not have to be defined at a point for it to have a limit there.
And that's the case for the example you gave.
 

use L`Hospital rule... limit is equal to limit of derivatives...
 

normally limits are used when the function is undefined at a particular point.
in ur case
1.since sin(x)/x will become undefined at x=0. the limits are taken
apply l hospital's rule
so cosx/1 ==> 1 at x=0
2. here also since 1-r --> 0 , but not exaclty 0, so we can cancel
 

smileysam said:
hello everyone...

secondly as r tends to 1 suppose we have the following equation (1-r^2)/(1-r)
then we cancel out (1-r) from numerator & denom ....whats the logic as (1-r) will tend
Thanks alot.
Click to expand...
This is always true:
(1+r)*(1-r)/(1-r)=1+r
and lim(r->1)= 2
 

Try the sandwich theorem for proving the limit sin(x)/x as x->0.

As x->0(for small values of x),sin(x)->x. Hence we are validated to the result 1.

Check out using a scientific calc for values of the sine of small angles.

As for the other limit, use the L'Hospitals rule defined for indeterminate forms(0/0 or ∞/∞).
 

thanks all of u.....
L hospital's rule is the right way to solve the prob.
 

You sure can use L'Hospital rule to find the limit for the problem.
But I wouldn't say " L'Hospital rule is the right way to solve it".
To understand the fundamental concept of limit, one should start with ''definition of limit".

rgds
 

Here is another method u can expand sinx in series

f(a+0)= f(0)+a.f'(0)+a.a.f''(0)/2...............
so for sinx=x+x^3/3!..........
so lim x-->0 sinx/x=lim x-->0 1+x^2/3!......= 1
 

For x>0, x->0:
sin(x) < x < tan(x)
thus,
sin(x)/x < x /x <1
sin(x)/x > x / tan(x) = x / sin(x) * cos(x)
sin(x)/x > sqrt(cos(x))
sqrt(cos(x)) -> 1 (when x -> 0+)
thus,
sin(x)/x -> 1 (when x-> 0+)

when x<0, x->0-:
sin(x)/x=sin(-x)/(-x) -> 1
 

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