# Finding the limit of a function with absolute values.

1. Mar 21, 2006

### 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?

Last edited: Mar 21, 2006
2. Mar 21, 2006

### e(ho0n3

Makes sense to me.

3. Mar 21, 2006

### 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.

4. Mar 21, 2006

### drunkenfool

Oh, how so? Am I doing this right?

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

5. Mar 22, 2006

### VietDao29

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:
$$\lim_{x \rightarrow 1 ^ -} \frac{x ^ 2 - 1 + |x - 1|}{|x - 1|}$$
$$= 1 + \lim_{x \rightarrow 1 ^ -} \frac{x ^ 2 - 1}{|x - 1|}$$
$$= 1 - \lim_{x \rightarrow 1 ^ -} \frac{(x - 1)(x + 1)}{x - 1}$$
$$= 1 - \lim_{x \rightarrow 1 ^ -} (x + 1)$$
$$= -1$$.

6. Mar 22, 2006

### 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.