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I tried to solve the follwing limit of function by delta and epsilon but I got stuck with finding the right "N" that for each $x<N$, $|f(x) - L|< \epsilon$, and the hard part was how to get x out of the absolute value, because $x$ is negative. can someone give a clear explanation of this point? and how do I complete this proof?

$\lim \limits_{x \to -\infty} \frac{9x-7x^2}{4x^2+8} = \frac{-7}{4}$

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  • $\begingroup$ The $N$ is needed when using the sequential definition of a limit, and at no point should $x$ be compared with $N$. You don't seem to have a sequence defined anywhere. Also, what is $f$? $\endgroup$ – graydad Dec 16 '14 at 15:17
  • $\begingroup$ well N is just name, i could have named it K. $f$ is a function. $\endgroup$ – Firas Ali Abdel Ghani Dec 16 '14 at 15:19
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It is not $x$ in the absolute value sign, it is $f(x)-L$. If it were $x$, it would be easy: $|x|=-x$ because $x \lt 0$. $f(x)-L=\frac {9x-7x^2}{4x^2+8}+\frac 74$ When you combine the fractions, the $x^2$ term will cancel in the numerator. The $x^2$ in the denominator will dominate, so if $x$ is large enough the error will be small.

Added: $f(x)-L=\frac {9x-7x^2}{4x^2+8}+\frac 74=\frac {9x-7x^2+7x^2+14}{4x^2+8}=\frac {9x+14}{4x^2+8}$ If $x \lt -2$ this will be negative, so $|f(x)-L|=\frac {14-9x}{4x^2+8}$ and the absolute signs are gone. You can make this arbitrarily small by making $x$ large and negative.

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  • $\begingroup$ I reached that $|\frac{9x+14}{8}|<\epsilon$ , then how can I isolate $x$? $\endgroup$ – Firas Ali Abdel Ghani Dec 16 '14 at 15:36
  • $\begingroup$ The denominator should be $4x^2+8$, not $8$, and the numerator is not right either. That $x^2$ is what makes the error small. $\endgroup$ – Ross Millikan Dec 16 '14 at 15:39
  • $\begingroup$ ok, I got it, but I still don't know how to say according to definition that for each $\epsilon$ there is a "value" that if x is smaller than this value, then everything we said is fulfilled. $\endgroup$ – Firas Ali Abdel Ghani Dec 16 '14 at 15:53
  • $\begingroup$ the first equation must be equal to $\frac{9x+14}{4x^2+8}$ $\endgroup$ – Firas Ali Abdel Ghani Dec 16 '14 at 17:16
  • $\begingroup$ You are right. Fixed. The next sentence was correct for this version $\endgroup$ – Ross Millikan Dec 16 '14 at 17:21
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Hint: $$ \left\lvert \frac{9x - 7x^2}{4x^2 + 8} + \frac{7}{4} \right\rvert = \left\lvert \frac{\frac{9}{x} - 7 + 7 \left( 1 + \frac{2}{x^2} \right)}{4 + \frac{8}{x^2}} \right\rvert = \left\lvert \frac{\frac{9}{x} + \frac{14}{x^2}}{4 + \frac{8}{x^2}} \right\rvert \le \frac{- 9 x}{4x^2 + 8} + \frac{14}{4x^2 + 8} $$ Now if $x$ is small enough (meaning if $x < -N$ for some large $N \in \mathbb{N}$) can we make the ratio $\frac{-x}{x^2}$ small? For instance can we make it smaller than say $\epsilon / 2$ for some $\epsilon$? Then what about $\frac{- 9 x}{4x^2 + 8}$? And can we make $\frac{1}{x^2}$ small? If so then what about $\frac{14}{4x^2 + 8}$?

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  • $\begingroup$ it should be $\frac{- 9 x}{4x^2 + 8} + \frac{14}{4x^2 + 8}$, and how can I say formally that each part of this is less than $\frac{\epsilon}{2}$? $\endgroup$ – Firas Ali Abdel Ghani Dec 16 '14 at 15:47
  • $\begingroup$ @FirasAliAbdelGhani ah yes, my apologies, forgot the factor of $7$. As for doing this precisely, how would you make just $-x / x^2$ smaller than $\epsilon / 2$? I would then extrapolate from here. I remember when doing these initial analysis proofs it seemed impossible to do so, but really doing this stuff yourself teaches you a wonderful amount of information, much more than I could ever possibly give you over MSE. $\endgroup$ – DanZimm Dec 16 '14 at 16:30

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