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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0that's quite long to type. haha. you can do it dude! :)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0sure i can ... wanted to jst chk

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0just plug 4 into the equation

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@suneja using epsilon and delta right

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@nickhouraney ya can u show dat here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0wats lim anyone pls help

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0xa x4 f(x)L<epsilon  (x^2+x11)  9 start from here

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0start with \[x^2x20<\epsilon\] and see what you need for \(x4\) to make it happen

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0mod (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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0we 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ya 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
 4 years ago
Best ResponseYou've already chosen the best response.0for 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0well lets see your proof

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0oh 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 \(x+5\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0trick 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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok 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 \(\delta\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you are looking at the product inside the square root, you have \((x4)(x+5)\) and you want to make this smaller than \(\epsilon\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0but \((x4)(x+5)=x4x+5\) you can make the first term as small as you like, but you cannot make \(x+5\) small

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0not square root, absolute value is what i meant

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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
 4 years ago
Best ResponseYou've already chosen the best response.0so you can assert that since \(x\) is close to 4, it is certainly less that say 5, making \(x+5<10\)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now you want \[(x4)(x+5)<\epsilon\] if you make \(\delta=\frac{\epsilon}{10}\) and \(3<x<5\) then you know \[(x4)(x+5)<10x4<10\times \frac{\epsilon}{10}=\epsilon\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[\infty \rightarrow +\infty\] is not a bound right?
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