1

I know that $\displaystyle \lim_{x \to \frac{3}{2}} \frac{2x^2-3x}{|2x-3|}$ does not exist, because the lateral limits are different and I also know that the absolute value on the denominator has something to do with it. But I can´t get my mind around it.

Can some one explain it to me slowly?

Seth
  • 9,393
  • 1
  • 27
  • 64
user3347814
  • 846
  • 2
    Do you mean $x \rightarrow \frac{3}{2}$? If so, make sure you also change the subject line. – Dave L. Renfro May 22 '14 at 20:22
  • 1
    Assuming you really mean to approach $\frac32$, the whole problem is that the numerator changes sign there but the denominator does not. So unless the limit is zero (and it isn't), you're in big trouble -- the left and right limits, if they exist, have opposite sign so can't be equal. – MPW May 22 '14 at 20:31
  • Sorry I wrote it wrong. I fixed it now... I meant $\frac{3}{2}$, my bad! – user3347814 May 22 '14 at 20:35

4 Answers4

3

Here's one way to think of it: Whenever you have a function defined using absolute values, you can replace it with a function defined in a piecewise manner: so,

$$f(x) = \frac{2x^2 - 3x}{|2x - 3|}$$

Is the same as:

$$f(x) = \left\{\begin{array}{lc} x \qquad & x > \frac{3}{2} \\ -x & x < \frac{3}{2} \end{array}\right.$$

By graphing this, you should be able to see why the limit does not exist.

Alex Zorn
  • 3,502
  • 15
  • 17
2

When taking the limit from the left, $2x-3$ is negative, so that you have $$ \lim_{x\to\frac{3}{2}^-} \frac{2x^2-3x}{3-2x} = \lim_{x\to\frac{3}{2}^-} (-x) = -\frac{3}{2}.$$ On the other hand, when taking the limit from the right, $2x-3$ is positive, so that the limit is $$ \lim_{x\to\frac{3}{2}^+} \frac{2x^2-3x}{2x-3} = \lim_{x\to\frac{3}{2}^+} x = \frac{3}{2}.$$ So the two one-sided limits are unequal.

rogerl
  • 22,399
2

Consider the graph of the following function:

$$\frac{x}{|x|}=\begin{cases} +1 & x>0 \\ -1 & x<0 \\ {\rm undefined} & x=0 \end{cases} $$

Obviously this doesn't have a limit at $0$.

Think about how this relates to the graph of $\displaystyle\frac{2x-3}{|2x-3|}\cdot x$.

anon
  • 151,657
2

The key is to realize that absolute value is a piecewise defined function in disguise. Recall that $|u| = u$ if $u > 0$ and $|u| = -u$ if $u < 0.$ In your case, $u = 2x – 3,$ so $u > 0$ corresponds to $x > \frac{3}{2}$ (solve $2x – 3 > 0)$ and $u < 0$ corresponds to $x < \frac{3}{2}$ (solve $2x – 3 < 0).$

Thus, we want to separately consider the right and left limits, making use of the fact that $|2x – 3|$ can be rewritten without the use of absolute value bars when we stay on one side of $\frac{3}{2}.$ Specifically, when we're on the right side of $\frac{3}{2}$ we have $2x – 3 > 0$ and hence $|2x – 3| = (2x – 3);$ and when we're on the left side of $\frac{3}{2}$ we have $2x – 3 < 0$ and hence $|2x – 3| = -(2x – 3).$

Therefore,

$$ \lim_{x \to {\frac{3}{2}}^{+}} \frac{x(2x – 3)}{|2x - 3|} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{+}} \frac{x(2x – 3)}{(2x – 3)} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{+}} \frac{x}{1} \;\; = \;\; \frac{3}{2} $$ and $$ \lim_{x \to {\frac{3}{2}}^{-}} \frac{x(2x – 3)}{|2x - 3|} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{-}} \frac{x(2x – 3)}{-(2x – 3)} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{-}} \frac{x}{-1} \;\; = \;\; -\frac{3}{2} $$

Dave L. Renfro
  • 36,843