Question

What is the limit-
lim x->-1
[1/sqrt(1+x)]-1
-------------
x

Answer 1

is L'Hopital's rule allowed(or have you been taught) ?

Answer 2

haven't learned it

Answer 3

I have heard about it a lot recently but I believe we are covering that next chapter

Answer 4

algebra to the rescue

Answer 5

okk...then lets do by conjugate method,

Answer 6

i have tried algebraically but i get 1/(sqrtx)+2x+x^2

Answer 7

which i cant answer because we cant really square root -1

Answer 8

\(\large \dfrac{1}{\sqrt{x+1}}-1=\dfrac{1-\sqrt{x+1}}{\sqrt{x+1}}\dfrac{1+\sqrt{x+1}}{1+\sqrt{x+1}}\)
have you done this only ?

Answer 9

then numeratorr will come out to be
1-(x+1)

Answer 10

i did that to get 1/(sqrt 1+x)(1+sqrt(1+x)

Answer 11

sorry i meant x/(sqrt 1+x)(1+(sqrt(1_x))

Answer 12

wait, why didn't u just substitute x=-1 ??

Answer 13

because it would give me a 0 in the denominator which is undefined

Answer 14

but im not too sure about it being undefined.

Answer 15

a/0 is not undefined
if a is positive, its +infinity
if a is negative, its -infinity

Answer 16

yes but because the two limits are not equal it is considered undefined or non existant

Answer 17

here, you need to find the left and right hand limits, you know what they are ?

Answer 18

the left side of the limit must equal the right side of the limit and -infinity doesn't equal +infinity

Answer 19

i know what the left and right sided limits are

Answer 20

yes, they come out to be unequal, and this limit actually DNE

Answer 21

ok well then what about this one?
lim x->-2
x^2/(x^2-4)

Answer 22

right hand limit = -infinity
left hand limit = i*infinity

Answer 23

i stated because we are approaching from the left we get negative infnity

Answer 24

for that left hand limit =+infinity
rihght hand limit =-infinity
hence that limit also DNE

Answer 25

could you think of how ?

Answer 26

i understand how, im just not sure about stating undefined in this question but i suppose it would work

Answer 27

so i guess for both answers it is negative infinity right?

Answer 28

no...both of these limits 'Does not exist'
because for both of these, left hand limit not = right hand limit.

Answer 29

negative infinity does not imply that limit DNE

Answer 30

yes the question is only asking for the left side of the limit and the left side of the limit does exist as negative infinity

Answer 31

ohh...but for 1st
LHL = i/infinity
for 2nd, LHL = +infinity

Answer 32

i thought for both it would be -infnity as it is asking x->-2 and x->-1

Answer 33

yes i understood that it is -infinity

Answer 34

let me explain the easier one,
when x->-2 from left means x is less than -2,but very near to it
so, x+2 is NEGATIVE but very near to 0
hence 1/(x+2) is
−∞ and there is (x-2) also, which is -4, which is negative!
so, overall, its +infinity.
got this ?

Answer 35

i am sorry, i typed the question wrong. For the easier one it is asking for the limit x->-2-

Answer 36

while for the other one it is simply asking for x->-1. So for the easier one it is asking for the left side of the limit while the harder one is just asking for the limit which we found to be DNE

Answer 37

yes, \(-2^-\)
means less than -2, but very near to it.
so, for 2nd its +infinity
and for 1st its DNE because LHL not =RHL

Answer 38

you got how +infinity ?

Answer 39

no i am still lost. I learned that −2 − or the left sided limit would mean negative infninty and the right sided limit would be poisitve infnity

Answer 40

i heard you saying you got how 1/(x+2) is -infinity, any doubts there ?

Answer 41

for this question-
[1/sqrt(1+x)]-1
-------------
x
I know it is DNE because the questions isn't asking for only one side of the limit

Answer 42

ok, so lets do that first,
to find that limit, we need to find both LHL and RHL
for LHL, we thake, \(\large -1^-\)
means x is very near to -1 but less than -1
for RHL, we thake, \(\large -1^+\)
means x is very near to -1 but greater than -1
got this much ?

Answer 43

yes

Answer 44

good, so when we find LHL and RHL, they don't come out to be =, and hence the original limit DNE.
right ? done with 1st ?

Answer 45

or you want to know how we found LHL and RHL ?

Answer 46

ok that i got now about the second question

Answer 47

to find 2nd limit, we again need to find both LHL and RHL
for LHL, we thake, \(−2^−\)
means x is very near to -2 but less than -2
for RHL, we thake, \(−2^+\)
means x is very near to -2 but greater than -2
can you find LHL and RHL on your own ?

Answer 48

yeah LHL is -infinity while RHL is +infinity

Answer 49

how ?

Answer 50

because you cant algebraically change the question

Answer 51

the question being x^2/x^2-4 with the lim x->-2

Answer 52

yes, the function is x^2/x^2-4 = x^2/(x+2)(x-2)
and we need to put x=-2
for LHL, \(x=-2^-\), so x+2 will be negative , right ?

Answer 53

ohhhh i see what i did wrong now

Answer 54

see if you get this :
x is less than -2,
so x+2 is negative, and near to 0
so, 1/(x+2) is -infinity

Answer 55

but we still have x-2 !

Answer 56

x^2/x^2-4 = x^2/(x+2)(x-2)
then we change it too-
-2^2/(-2+2)(-2-2)
which would give is -
4/-4

Answer 57

no it wont
my bad it would still give use 0 h

Answer 58

and just in case, the limit isn't asking for -2- it is only -2

Answer 59

unless you are clear with this,
1/(x+2) is -infinity
lets not proceed....

Answer 60

that i understood if it was asking for the LSL but its not asking for the LSL

Answer 61

even if its only -2, we will need to find both LHL(-2-) and RHL(-2+)

Answer 62

yes, so,
1/(x+2) is -infinity
but x-2 = -4 is also negative!
so, negative from -4 and -infinity combines to give you +infinity.

Answer 63

yes we do, or we can simply plug in 2 itself and find thenswer

Answer 64

you meant -2 ? no....we need to find both LHL and RHL....
if they are unequal, limit DNE, as in this case.
if you only plug in -, you'll get +infinity, which is incorrect.

Answer 65

**only plug in -2,

Answer 66

yeah my bad i meant -2 and i see what you mean but we would only need to do that if we had two separate functions. We only have one function here and so we only need to plug in one number

Answer 67

in other words the left sided limit begins at -2 while the right also begins at -2 so there is no need for two equations.

Answer 68

there's only one equation, x^2/(x+2)(x-2)
but when we take LHL / RHL, we plug in different values, one less than -2, one greater than -2
thats what they mean, if the curve approaches the same value from left and right , only then limit exist, which is not case here,
x^2/(x+2)(x-2) approaches +infinity from left
and -infinity from right.

Answer 69

**from left of -2
**from right of -2

Answer 70

and it does approach the same value doesn't it?

Answer 71

i mean if we plug in -1.99999999999 and -2.00000000001 into x we would get almost the same numbers

Answer 72

http://www.wolframalpha.com/input/?i=x%5E2%2F%28x%2B2%29%28x-2%29
see the curve at x=-2
from left it goes UP, to +infinity
from right , it goes DOWN to -infinity
so, no,it does not approach the same value