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lim x^2+x11=9
x tends to 4
prove tis by formal definition of limit
 one year ago
 one year ago
lim x^2+x11=9 x tends to 4 prove tis by formal definition of limit
 one year ago
 one year ago

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Jeffrey_CalderonBest ResponseYou've already chosen the best response.0
that's quite long to type. haha. you can do it dude! :)
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
sure i can ... wanted to jst chk
 one year ago

heedcomBest ResponseYou've already chosen the best response.0
just plug 4 into the equation
 one year ago

nickhouraneyBest ResponseYou've already chosen the best response.1
@suneja using epsilon and delta right
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
@nickhouraney ya can u show dat here
 one year ago

rerukumashBest ResponseYou've already chosen the best response.0
wats lim anyone pls help
 one year ago

rerukumashBest ResponseYou've already chosen the best response.0
ok where is it used
 one year ago

nickhouraneyBest ResponseYou've already chosen the best response.1
xa x4 f(x)L<epsilon  (x^2+x11)  9 start from here
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
start with \[x^2x20<\epsilon\] and see what you need for \(x4\) to make it happen
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
mod (x^2x11)9 < epsilon wenever 0<mod x4 < 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
 one year ago

nickhouraneyBest ResponseYou've already chosen the best response.1
we basically need to take this  (x^2+x11)  9 < epsilon and make it x4<epsilon start by simplifying  (x^2+x11)  9 that will turn to  x^2+x11  9 < epsilon
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
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
 one year ago

nickhouraneyBest ResponseYou've already chosen the best response.1
for the proof you would just go in reverse basically. if x4 < delta then x4 < (your choice for epsilon) then show how this will lead to f(x)L < epsilon
 one year ago

nickhouraneyBest ResponseYou've already chosen the best response.1
well lets see your proof
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
i've done exactly like but was lil confused in 2nd half of it! neways thanks
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
oh yes it is. okay you should start with \[x^2+x20<\epsilon\] then factor to get \[(x4)(x+5)<\epsilon\] you have control over \(x4\) so you just need a bound for \(x+5\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
wat do u mean by " a bound for \[\left x+5 \right\]
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
ok lets go slow you get to say how large \(x4\) can be, that is you get to pick your \(\delta\) so that \(x4\) is smaller than \(\delta\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now you are looking at the product inside the square root, you have \((x4)(x+5)\) and you want to make this smaller than \(\epsilon\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
but \((x4)(x+5)=x4x+5\) you can make the first term as small as you like, but you cannot make \(x+5\) small
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
not square root, absolute value is what i meant
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
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
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
so you can assert that since \(x\) is close to 4, it is certainly less that say 5, making \(x+5<10\)
 one year ago

satellite73Best ResponseYou've already chosen the best response.1
now you want \[(x4)(x+5)<\epsilon\] if you make \(\delta=\frac{\epsilon}{10}\) and \(3<x<5\) then you know \[(x4)(x+5)<10x4<10\times \frac{\epsilon}{10}=\epsilon\]
 one year ago

sunejaBest ResponseYou've already chosen the best response.0
\[\infty \rightarrow +\infty\] is not a bound right?
 one year ago
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