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

sure i can ... wanted to jst chk

just plug 4 into the equation

@nickhouraney ya
can u show dat here

wats lim anyone pls help

lim is limit

ya i can show u

ok where is it used

pls reply

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

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

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

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

@nickhouraney thanks

well lets see your proof

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

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

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

not square root, absolute value is what i meant

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

got it ..thanks all

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