Finding the limit of a function with absolute values.

[drunkenfool]

I need a little help and reassurance here.

The question is as follows,

Find the following limit, if it exists.

$$\lim_{x \rightarrow 1} \frac{x ^ 2 + |x -1| - 1}{|x-1|}$$

Here is what I did, first I did the two one-sided limits, as $$\lim_{x \rightarrow 1^+}$$ and as $$\lim_{x \rightarrow 1^-}$$. (the values in the absolute value would be (x-1) and -(x-1) respectively, in this case) The answers I got were 3 and -1 respectively. Since the one-sided limits aren't the same, I concluded that the limit for this function does not exist. Am I correct?

 

[e(ho0n3]

Makes sense to me.

 

[HallsofIvy]

One correction: the limit as x approaches 1 from below, $$\lim_{x \rightarrow 1^-}$$, is 1, not -1. Of course, the limit still doesn't exist.

 

[drunkenfool]

Oh, how so? Am I doing this right?

=$$\frac{x ^ 2 + |x -1| - 1}{|x-1|}$$
=$$\frac{x ^ 2 + -(x -1) - 1}{-(x-1)}$$
=$$\frac{x ^ 2 -x}{-(x-1)}$$
=$$\frac{-x (1-x)}{1-x}$$
=$$-x$$
=$$-1$$

 

[VietDao29]

Oh, how so? Am I doing this right?

=$$\frac{x ^ 2 + |x -1| - 1}{|x-1|}$$
=$$\frac{x ^ 2 + -(x -1) - 1}{-(x-1)}$$
=$$\frac{x ^ 2 -x}{-(x-1)}$$
=$$\frac{-x (1-x)}{1-x}$$
=$$-x$$
=$$-1$$

Yes, -1 is the correct answer. However, it's not very clear the way you wrote it. Why are your equal signs all fly off to the numerator, and where are all the lim notation?
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Or you may try to get rid of the |x - 1| in the numerator first, and then apply the limit:
$$\lim_{x \rightarrow 1 ^ -} \frac{x ^ 2 - 1 + |x - 1|}{|x - 1|}$$
$$= 1 + \lim_{x \rightarrow 1 ^ -} \frac{x ^ 2 - 1}{|x - 1|}$$
$$= 1 - \lim_{x \rightarrow 1 ^ -} \frac{(x - 1)(x + 1)}{x - 1}$$
$$= 1 - \lim_{x \rightarrow 1 ^ -} (x + 1)$$
$$= -1$$.

 

[drunkenfool]

Oh, I really don't know my way around the latex codes, so that's why you see all the errors. Thanks a lot, you guys.