Question

lim x^2+x-11=9
x tends to 4
prove tis by formal definition of limit

Answer 1

that's quite long to type. haha. you can do it dude! :)

Answer 2

sure i can ... wanted to jst chk

Answer 3

just plug 4 into the equation

Answer 4

@suneja using epsilon and delta right

Answer 5

@nickhouraney ya
can u show dat here

Answer 6

wats lim anyone pls help

Answer 7

lim is limit

Answer 8

ya i can show u

Answer 9

ok where is it used

Answer 10

pls reply

Answer 11

|x-a|
|x-4|

|f(x)-L|<epsilon
| (x^2+x-11) - 9|

start from here

Answer 12

start with
\[|x^2-x-20|<\epsilon\] and see what you need for \(|x-4|\) to make it happen

Answer 13

mod (x^2-x-11)-9 < epsilon wenever 0<mod x-4 < delta
tis is how v start
then go by simplifying left inequality to find a suitable value for delta
then verify that choice of delta
i hav a prob in verification .. need help der

Answer 14

we basically need to take this | (x^2+x-11) - 9| < epsilon and make it |x-4|<epsilon

start by simplifying | (x^2+x-11) - 9|
that will turn to | x^2+x-11 - 9| < epsilon

Answer 15

ya
then u hav to prove that "choice " of delta wic u get by solving the 1st part like u said
can u show me 2nd part of tis prob

Answer 16

for the proof you would just go in reverse basically.

if |x-4| < delta
then |x-4| < (your choice for epsilon)
then show how this will lead to |f(x)-L| < epsilon

Answer 17

@nickhouraney thanks

Answer 18

well lets see your proof

Answer 19

i've done exactly like but was lil confused in 2nd half of it! neways thanks

Answer 20

oh yes it is. okay you should start with
\[|x^2+x-20|<\epsilon\] then factor to get
\[|(x-4)(x+5)|<\epsilon\] you have control over \(|x-4|\) so you just need a bound for \(|x+5|\)

Answer 21

trick is to say that since you are taking the limit as \(x\to 4\) you can assert that say \(3<x<5\) so the largest \(|x+5|\) can be is 10

Answer 22

wat do u mean by " a bound for \[\left| x+5 \right|\]

Answer 23

ok lets go slow
you get to say how large \(|x-4|\) can be, that is you get to pick your \(\delta\) so that \(|x-4|\) is smaller than \(\delta\)

Answer 24

now you are looking at the product inside the square root, you have
\(|(x-4)(x+5)|\) and you want to make this smaller than \(\epsilon\)

Answer 25

but
\(|(x-4)(x+5)|=|x-4||x+5|\)
you can make the first term as small as you like, but you cannot make \(|x+5|\) small

Answer 26

not square root, absolute value is what i meant

Answer 27

so the question is, how big can \(|x+5|\) be? well it can be really really large, but don't forget you are making \(x\) close to 4

Answer 28

so you can assert that since \(x\) is close to 4, it is certainly less that say 5, making \(|x+5|<10\)

Answer 29

now you want
\[|(x-4)(x+5)|<\epsilon\] if you make
\(\delta=\frac{\epsilon}{10}\) and \(3<x<5\) then you know
\[|(x-4)(x+5)|<10|x-4|<10\times \frac{\epsilon}{10}=\epsilon\]

Answer 30

got it ..thanks all

Answer 31

\[-\infty \rightarrow +\infty\] is not a bound right?