3
msp
·2009-02-04 08:26:54
i can give u an example where the sandwich theorem is applied
x-1≤[x]≤x
[] shows GINT
106
Asish Mahapatra
·2009-02-04 07:53:00
i never said anything abt continuity sky.
BTW, can anyone explain me sandwich theorem?
1
skygirl
·2009-02-04 07:58:07
lim (e^1/x - 1)/(e^1/x + 1)
x-->0
∞/∞ form.
use L'hospital.
limit=1.
------------------------------------------------------------------------------------------------
by traditional method,
LHL:
x-0 = -h.
lim h->0 [(e^1/h - 1)/(1/h)]/[(e^1/h - 1 + 2)/(1/h)]
=1. [lim x->0 (e^x-1)/x =1 ]
similarly RHL will also be 1.
so limit=1.
1
skygirl
·2009-02-04 07:59:03
@asish,,, u din tell!! i told akand... he was telling something about comtinuity...
106
Asish Mahapatra
·2009-02-04 08:00:47
@sky: this question was in RD Sharma
LHL is not x->0-
WHY?
62
Lokesh Verma
·2009-02-04 08:10:33
lim (e^1/x - 1)/(e^1/x + 1)
x-->0+
e1/x>>1
hence the limit will be 1
when
lim (e^1/x - 1)/(e^1/x + 1)
x-->0-
e1/x will be close to 0
hence limit will be -1
SO RHL=1
LHL=-1
Hence the limit does not exists
1
Akand
·2009-02-04 08:14:44
coz u made a mistake sumwher..........a silly one i suppose...........k wats the answer 4 d given ques??? limit doesnt exist eh?
3
msp
·2009-02-04 08:18:33
try this link da http://en.wikipedia.org/wiki/Squeeze_theorem
1
skygirl
·2009-02-04 07:52:28
sky, if u take RHL then
x-->0_ ==> e^1/x = e^-∞ = 0
then limit = -1/1 = -1
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LHL is not x->0-
is it?
it is jus checking the limit at the nearby region of x=0... but on the left hand side.
62
Lokesh Verma
·2009-02-04 08:44:22
limit x tending to infinity
[x]/x
1
RAY
·2009-02-04 09:03:35
whts d sandwitch theorum????
doesnt seem like sumthing good to eat......
1
nihal raj
·2010-10-01 20:31:02
limit exists..... by taking common frm num. and den.
1
sahil jain
·2010-10-01 22:02:17
limit exists and answer is 1
3
Abhishek Majumdar
·2010-10-15 23:07:56
1) Lt {x}/tan{x}
x→0
Solution....
RHL=Lt x-[x]/tan(x-[x]) =Lt x/tanx....as when x→o+,[x]→0 !!
x→0+ x→0+
thus RHL=1
LHL=Lt x-[x]/tan(x-[x])=Lt x+1/tan(x+1)....since when x→0-,[x]→--1
x→0- x→0-
thus LHL=Cot1
thus LHL≠RHL...THUS LIMIT DOEST NT EXIST !!
1
sahil jain
·2010-11-18 10:18:59
hey man , limit doesnt exist
1
souradeepmajumde majumder
·2010-11-19 04:49:19
for first question limit does not exixt........
because in rhl you find 1............
but in lhl you will find

you can use box0-h when h-o is -h-1.......... you can obtain this result........
3
msp
·2009-02-04 07:39:35
in this ques the limit doesnt exist
1
Akand
·2009-02-04 07:32:17
because {0+h} is h so..........h/tanh is 1..... i suppose
1
Akand
·2009-02-04 07:32:42
but when u take LHL.........we are dead......
1
Akand
·2009-02-04 07:33:31
{0-h} is not h............so LHL not eqaul to RHL..............im not sure
106
Asish Mahapatra
·2009-02-04 07:33:41
yes i meant that if u evaluate using LHL and RHL then RHL = 1
but LHL is not 1 hence i think limit wont exist
1
skygirl
·2009-02-04 07:36:03
yes sankara ... i din c ur post ...
106
Asish Mahapatra
·2009-02-04 07:36:50
see another one:
lim (e^1/x - 1)/(e^1/x + 1)
x-->0
If you divide e^1/x throughout and then evaluate limit, u will get the answer as 1 but if u take LHL and RHL then LHL=1 and RHL=-1.
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?