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suneja
lim x^2+x-11=9 x tends to 4 prove tis by formal definition of limit
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
@suneja using epsilon and delta right
@nickhouraney ya can u show dat here
wats lim anyone pls help
|x-a| |x-4| |f(x)-L|<epsilon | (x^2+x-11) - 9| start from here
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<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
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
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
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
well lets see your proof
i've done exactly like but was lil confused in 2nd half of it! neways thanks
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 \(|x+5|\)
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
wat do u mean by " a bound for \[\left| x+5 \right|\]
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 \(\delta\)
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 \(\epsilon\)
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
not square root, absolute value is what i meant
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
so you can assert that since \(x\) is close to 4, it is certainly less that say 5, making \(|x+5|<10\)
now you want \[|(x-4)(x+5)|<\epsilon\] if you make \(\delta=\frac{\epsilon}{10}\) and \(3<x<5\) then you know \[|(x-4)(x+5)|<10|x-4|<10\times \frac{\epsilon}{10}=\epsilon\]
\[-\infty \rightarrow +\infty\] is not a bound right?