Question

confusion with limits
lim of (sin(x))/(x) as x approaches 0 = 1
but y is there no limit for lim of (1/(x^2)) as x approaches 0?

Answer 1

since the reason is that if we put 0 in the limit directly the numerator gets to 0 in sinx/x but in the latter it will turn into 1/0 form which is undetermined........

Answer 2

ye but sin of z equals 1

Answer 3

wait nvm

Answer 4

didnt get u???

Answer 5

ye but if the denominator is 0 then shouldn't sinx /x also be undefined?

Answer 6

first in case of sin we have numerator to be taken first na?

Answer 7

ye so ?

Answer 8

http://www.youtube.com/watch?v=Ve99biD1KtA&feature=player_embedded

Answer 9

try this if u can have patience to see... :) :) :) this is the proof for your doubt about sinx/x actually its a proved hypothesis... but fails in case of 1/x^2

Answer 10

so here is my question

Answer 11

u can write limts as x is approaching some asymptote right?

Answer 12

if that is true then y is that finction along with these functions also undefined
lim of ((4+x)^2-16)/x as x approaches 0

Answer 13

or lim of abs(x)/x as x approaches 0

Answer 14

all three of these functions are undefined

Answer 15

i understand y the abs(x)/x is undefined but

Answer 16

the 4+x)^2+16/x is a hole

Answer 17

so then it still has a continous limit just a hole at that point

Answer 18

same with 1/x^2 so x=0 is a limt

Answer 19

i mean asymptote

Answer 20

there are some proofs on these dude... see actually limits are governed by some rules and formulae you have to think how to apply them, not to think how they have come.... I cant tell you rather than that because, no one would ever think off

Answer 21

so there should be a asymptotic limit

Answer 22

yes.............

Answer 23

so y isnt there?

Answer 24

it should matter if the denominator is 0 and undefined becasue a limit only states that as x approaches 0 not when x is zero

Answer 25

it shouldn't *

Answer 26

@Curry , yes that is hole so its value is indeterminate or 0/0 but the limit as you approach the hole may have a real value...in fact it is 8

lets look at the graph of sin(x)/x
notice its approaching 1 at x=0

Answer 27

so this function has a limit even though it's denominator is 0.
lim of (x^2-3x+2)/(x^2-4)

Answer 28

so the book says there is no limit for the hole

Answer 29

http://www.wolframalpha.com/input/?i=plot+sin%28x%29%2Fx+from+-pi%2F2+to+pi%2F2

Answer 30

i understand the sinx / x

Answer 31

y don't the other functions have one

Answer 32

but the last function i typed has a limit so how is that possible even though the denominator for that funciton would also be 0 but for the other functions i typed the denominator 0 but limit is undefined??

Answer 33

that last function is 0/0 when x=2 , this means there could be a limit
when its n/0 then limit is undefined or you could say there is an asymptote

Answer 34

if you factor it you get
(x-2)(x-1) / (x-2)(x+2) = (x-1)/(x+2)

at x=2, it equals 1/4 so the limit is 1/4 even though technically there is a hole

Answer 35

exactly so the (4+x)^2-16/x also has a hole

Answer 36

but y doesn't that have a limt

Answer 37

anyway, the point is when you evaluate a limit and you get 0/0 , that just means you don;t know what the limit is...it doesn't mean its undefined or doesn't exist
you don;t have enough info

Answer 38

but waht is it isn't 0/0 its n/0 then still u can have limit right

Answer 39

it does have a limit at y=8
http://www.wolframalpha.com/input/?i=plot+%28%28x%2B4%29^2+-16%29%2Fx+from+-2+to+2

Answer 40

but my book says no limit

Answer 41

to find the limit algebraically, you would have to change the form of the function so that you don't get 0/0 when you plug in x=0
\[\frac{(x+4)^{2} -16}{x} = \frac{x^{2} +8x}{x} = x+8\]

Answer 42

exactly

Answer 43

thats what i thought and i put the limit as 8 as x approaches 0

Answer 44

but it says thats wrong

Answer 45

it says its undefined

Answer 46

umm i guess i would disagree with your book...are you sure its talking about the limit or the value
technically, that value at x=0 is undefined

Answer 47

the direction was find the limit if it exists

Answer 48

http://www.wolframalpha.com/input/?i=lim+%28%28x%2B4%29^2+-16%29%2Fx+as+x-%3E0

Answer 49

and the problem was (4+x)^2-16)/x

Answer 50

so then the limit for (1/x^2) as x approaches 0 is +infin right or would i be wrong to enter that on a test

Answer 51

yes you are correct...limit = pos infinity

Answer 52

any number divided by 0 will give you limit of either +-infinity

Answer 53

ok so im not wrong right? here was my source my book doesn't have the answers for the evens so i used the online solution here is the link

Answer 54

http://www.slader.com/textbook/9780201324457-calculus-graphical-numeric-algebraic-finney-demana-et-al-3rd-edition/62/


they were problems 18 and 20

Answer 55

?

Answer 56

hmm not sure, looks like maybe they havn't got into infinity or special 0/0 limits yet and they accept it as limit does not exist at this point
--i can't see their entire solution or explanation

but full correct answers are +infinity and 8

Answer 57

this is a calculus course correct?

Answer 58

calc C