Question

Limits help
lim x->i- e^{x}-{x}-1/{x}^2

Answer 1

\[\Huge \lim_{x \rightarrow I^{-}} \frac{e^{(x)}-{(x)}-I}{{(x)^2}}\]where (x)=FRACTIONAL PART OF X.

Answer 2

what is I?

Answer 3

any Integer

Answer 4

i belongs to interger, then fractional part of it should be 0.

Answer 5

yup.

Answer 6

that is why we have limit :O 1/0

Answer 7

yea, so the limit is infinity.

Answer 8

i have the options

Answer 9

1/2
e-2
I
does not exist

Answer 10

does not exist

Answer 11

it says TENDING to integer not exact integer..

Answer 12

where did you find this question?

Answer 13

textbook

Answer 14

it cannot be determined.
i guess, terms cannot be adjusted, in order to calculate the limit.

Answer 15

\[\Huge \lim_{x \rightarrow I^{-}} \frac{e^{(x)}-{(x)}-1}{{(x)^2}}\]

Answer 16

how about now :o

Answer 17

is it e-2?

Answer 18

u may let \(t=(x)\) it becomes\[ \lim_{t \rightarrow 0} \frac{e^t-t-1}{{t^2}}\]u can use series expansion of \(e^t\) or LHR

Answer 19

so..1/2?

Answer 20

1/2 what i got as answer :)

Answer 21

thankyou! did not think of this anyways. nice!

Answer 22

if (x) is the fractional part of a number then we would have for example (5.9)=.9 ...correct?

Answer 23

yes.

Answer 24

so what if I was 6...then you would be coming from the lft...so numbers like

5.9
5.99
5.999



so as \(x\to 6\) we have \((x)\to 1\)

Answer 25

from the left

Answer 26

thats right, we made a mistake

Answer 27

then again...what if I was -6

Answer 28

x-->6, {x}-->0
{x} belongs to [0,1).

Answer 29

if -6 then (x) --> 0

Answer 30

i guess not.

Answer 31

so the limit depends on \(I\)

Answer 32

any real number t can be express as
\[t=I+f\]
where I belongs to an interger
and f is called the fractional part of t where
\[f \epsilon [0,1)\]

Answer 33

well thats right\[\text{{x}=}x-[x]\]and \(\text{{x}} \in [0,1)\)

Answer 34

then how can {x} be equal to 1?

Answer 35

it approaches to 1 and is not equal to 1... thats what we need in taking limit :)

Answer 36

a limit point need not be a member of the original set

Answer 37

then there exists 2 limits
the left and the right limit.

Answer 38

a right hand limit would be more interesting

Answer 39

previous one was the left limit.
what about the right limit then?

Answer 40

the right would be the 1/2 you previously obtained.

Answer 41

exactly..botht the limits are not equal.

Answer 42

the limit from the left is not the same as from the right

Answer 43

then, limit at x=I does not exist.?

Answer 44

this

\[\lim_{x \to I} \frac{e^{(x)}-{(x)}-1}{{(x)^2}}\]

does not exist

Answer 45

does not exist.

Answer 46

if you are allowed to come from both direction then the limit does not exist....if you just come from the right then the answer is 1/2

Answer 47

yes.

Answer 48

you should also double check that you wrote down the problem correctly.

Answer 49

i didn't give this question.

Answer 50

well, now i see, its given limit at the left., it was not visible..lol!
i didn't see this. not visible properly.

Answer 51

ah...I guess we lost the original poster a while back ;)

Answer 52

eh yea..that was silly, but the discussion was good.