# Can somebody explain the resolution of the following limit? [closed]

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

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 Cookie, Claude Leibovici, user88595, Hakim, Eric StuckyJun 21 '14 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 '14 at 9:05

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