limits

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

x-1≤[x]≤x

[] shows GINT

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.

then y am i getting 1 ?

[3]

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

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 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 !!

hey man , limit doesnt exist

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........

because {0+h} is h so..........h/tanh is 1..... i suppose

but when u take LHL.........we are dead......

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?