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?
$\lim \limits_{x \to -\infty} \frac{9x-7x^2}{4x^2+8} = \frac{-7}{4}$
On
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 $\frac{14}{4x^2 + 8}$?
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.