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$$\lim_{x\to-\infty}\frac{x-1}{|1-x|}$$

I checked at wolfram alpha and the result is $-1$, but i can't understand the resolution yet.

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closed as off-topic by le gâteau au fromage, Claude Leibovici, user88595, Hakim, Eric Stucky Jun 21 at 11:00

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Why do you not understand? What is your reasoning and what do you think the solution is? –  user88595 Jun 21 at 9:05

1 Answer 1

up vote 4 down vote accepted

For all $ x \leq 1 $, we have $ |1 - x| = 1 - x $. Therefore, $$ \lim_{x \to - \infty} \frac{x - 1}{|1 - x|} = \lim_{x \to - \infty} \frac{x - 1}{1 - x} = \lim_{x \to - \infty} \frac{x}{- x} = -1. $$


Additional Information

I just realized that my original answer is unnecessarily complicated (as I typed it in a hurry). It can be simplified as follows.

For all $ x \leq 1 $, we have $ |1 - x| = 1 - x $. Hence, $$ \forall x < 1: \quad \frac{x - 1}{|1 - x|} = \frac{x - 1}{1 - x} = \frac{x - 1}{- (x - 1)} = -1. $$ Therefore, $$ \lim_{x \to - \infty} \frac{x - 1}{|1 - x|} = \lim_{x \to - \infty} -1 = -1. $$

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