<s>[INST] Calculate $\displaystyle \lim_{x \to \frac{3}{2}} \frac{2x^2-3x}{|2x-3|}$ 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?
 [/INST] 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
$$ \lim_{x \to {\frac{3}{2}}^{-}} \frac{x(2x – 3)}{|2x - 3|} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{-}} \frac{x(2x – 3)}{-(2x – 3)} \;\; = \;\; \lim_{x \to {\frac{3}{2}}^{-}} \frac{x}{-1} \;\; = \;\; -\frac{3}{2} $$
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