106
Asish Mahapatra
·2009-02-04 07:41:11
luk at the question
We are taught to just put the value first to evaluate the limit, the limit then if its indeterminate then to simplify it..but here, if we do so, answer is wrong!!!! so how do we do it?
will we check LHL and RHL of each and every question?
3
msp
·2009-02-04 07:44:44
yes asish we have to check LHL and RHL becos for a function to exist we have to get RHL=LHL
1
skygirl
·2009-02-04 07:45:03
how are u getting -1 and 1 ??
i am getting 1 in every case !
1
Akand
·2009-02-04 07:46:08
hey sanky...........we can check for LHL=RHL for limits to exists, and also for continuity also...........
106
Asish Mahapatra
·2009-02-04 07:48:10
sky, if u take LHL then
x-->0_ ==> e^1/x = e^-∞ = 0
then LHL = -1/1 = -1
1
skygirl
·2009-02-04 07:48:24
another imp thing ashish :
for a limit to exist it is never necessary for the function to be continuous or have the same value as RHL and LHL .
i mean if for lim x->a f(x), LHL=RHL=b and f(a)=p ..
then also we will say .. limit exists .
1
skygirl
·2009-02-04 07:49:56
@ akand, for continuity , RHL= LHL = value at the limiting point.
1
Akand
·2009-02-04 07:52:02
ya skygirl...........sorry left tht