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Prove: $$\lim_{x\, \to \,-\infty}⁡ \dfrac{ x^2+2x-3 }{ x^2+1 }=1$$

While making evaluations on my draft, i get: $|f(x)-L|= \dfrac{2}{|x-1|}$

I want to "remove" the absolut value in order to find epsilon. What kind of evaluations am i alowed to do? Taking into consideration that $m < 0$ and $x < m$.

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  • $\begingroup$ Remember, the limit of the division is the division of the limits: $\lim \frac{a}{b}=\frac{\lim a}{\lim b}$. Thus, when you evaluate the limits at $\infty$, you can drop the lower-order terms out of the equation: $\lim \frac{x^2+2x-3}{x^2+1}\rightarrow \lim \frac{x^2}{x^2} = \lim 1$. $\endgroup$ – FundThmCalculus Nov 15 '14 at 11:10
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Here's one way to remove the absolute values:

For $x<0$, we have $\left|\frac{x^2+2x-3}{x^2+1}-1\right|=\left|\frac{2x-4}{x^2+1} \right|=\frac{-2x-4}{x^2+1}$

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