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suneja
Group Title
lim x^2+x11=9
x tends to 4
prove tis by formal definition of limit
 2 years ago
 2 years ago
suneja Group Title
lim x^2+x11=9 x tends to 4 prove tis by formal definition of limit
 2 years ago
 2 years ago

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Jeffrey_Calderon Group TitleBest ResponseYou've already chosen the best response.0
that's quite long to type. haha. you can do it dude! :)
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
sure i can ... wanted to jst chk
 2 years ago

heedcom Group TitleBest ResponseYou've already chosen the best response.0
just plug 4 into the equation
 2 years ago

nickhouraney Group TitleBest ResponseYou've already chosen the best response.1
@suneja using epsilon and delta right
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
@nickhouraney ya can u show dat here
 2 years ago

rerukumash Group TitleBest ResponseYou've already chosen the best response.0
wats lim anyone pls help
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
lim is limit
 2 years ago

nickhouraney Group TitleBest ResponseYou've already chosen the best response.1
ya i can show u
 2 years ago

rerukumash Group TitleBest ResponseYou've already chosen the best response.0
ok where is it used
 2 years ago

rerukumash Group TitleBest ResponseYou've already chosen the best response.0
pls reply
 2 years ago

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

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
start with \[x^2x20<\epsilon\] and see what you need for \(x4\) to make it happen
 2 years ago

suneja Group TitleBest 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
 2 years ago

nickhouraney Group TitleBest 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
 2 years ago

suneja Group TitleBest 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
 2 years ago

nickhouraney Group TitleBest 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
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
@nickhouraney thanks
 2 years ago

nickhouraney Group TitleBest ResponseYou've already chosen the best response.1
well lets see your proof
 2 years ago

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

satellite73 Group TitleBest 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 \(x+5\)
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
wat do u mean by " a bound for \[\left x+5 \right\]
 2 years ago

satellite73 Group TitleBest 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 \(\delta\)
 2 years ago

satellite73 Group TitleBest 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 \(\epsilon\)
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest ResponseYou've already chosen the best response.1
not square root, absolute value is what i meant
 2 years ago

satellite73 Group TitleBest 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
 2 years ago

satellite73 Group TitleBest 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 \(x+5<10\)
 2 years ago

satellite73 Group TitleBest 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 \[(x4)(x+5)<10x4<10\times \frac{\epsilon}{10}=\epsilon\]
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
got it ..thanks all
 2 years ago

suneja Group TitleBest ResponseYou've already chosen the best response.0
\[\infty \rightarrow +\infty\] is not a bound right?
 2 years ago
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