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

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- anonymous

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

- schrodinger

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- anonymous

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

- anonymous

sure i can ... wanted to jst chk

- anonymous

just plug 4 into the equation

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## More answers

- anonymous

@suneja using epsilon and delta right

- anonymous

@nickhouraney ya
can u show dat here

- anonymous

wats lim anyone pls help

- anonymous

lim is limit

- anonymous

ya i can show u

- anonymous

ok where is it used

- anonymous

pls reply

- anonymous

|x-a|
|x-4|
|f(x)-L|

- anonymous

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

- anonymous

mod (x^2-x-11)-9 < epsilon wenever 0

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

@nickhouraney thanks

- anonymous

well lets see your proof

- anonymous

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

- anonymous

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 \(|x+5|\)

- anonymous

trick is to say that since you are taking the limit as \(x\to 4\) you can assert that say \(3

- anonymous

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

- anonymous

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 \(\delta\)

- anonymous

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 \(\epsilon\)

- anonymous

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

- anonymous

not square root, absolute value is what i meant

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

got it ..thanks all

- anonymous

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

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