limits

i can give u an example where the sandwich theorem is applied

x-1≤[x]≤x

[] shows GINT

i never said anything abt continuity sky.

BTW, can anyone explain me sandwich theorem?

lim (e^1/x - 1)/(e^1/x + 1)
x-->0

∞/∞ form.

use L'hospital.

limit=1.

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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.

@asish,,, u din tell!! i told akand... he was telling something about comtinuity...

@sky: this question was in RD Sharma
LHL is not x->0-
WHY?

coz u made a mistake sumwher..........a silly one i suppose...........k wats the answer 4 d given ques??? limit doesnt exist eh?

try this link da http://en.wikipedia.org/wiki/Squeeze_theorem

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.

whts d sandwitch theorum????

doesnt seem like sumthing good to eat......

limit exists..... by taking common frm num. and den.

limit exists and answer is 1

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 !!

LIMIT DOES NOT EXISTS.

in this ques the limit doesnt exist

how?

{0-h} is not h............so LHL not eqaul to RHL..............im not sure

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

will it exist [7]

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?