Evaluate the limit of (2x^2 + x - 3)/|x - 1| as x approaches 1, or show that it does not exist.

Step-by-step solution

1. 1

   Confirm the indeterminate form. At $x=1$ the numerator is $2+1-3=0$ and the denominator is $|0|=0$, so the expression is $\tfrac00$ and substitution gives nothing. Unlike the same numerator evaluated at $x=-2$, here the limit point is the zero of the absolute value, so the answer will be qualitatively different.

2. 2

   Factor the numerator by splitting the middle term. For $2x^{2}+x-3$ look for two numbers multiplying to $2\times(-3)=-6$ and adding to $1$: they are $3$ and $-2$. Then

   $2x^{2}+3x-2x-3=x(2x+3)-1(2x+3)=(x-1)(2x+3).$

   The factor $(x-1)$ matches what is inside the absolute value, which is exactly what makes the one-sided analysis necessary.

3. 3

   Take the left-hand limit. For $x<1$, $x-1<0$ so $|x-1|=-(x-1)$ and

   $\frac{(x-1)(2x+3)}{-(x-1)}=-(2x+3)\ \longrightarrow\ -(2+3)=-5.$

   So $\displaystyle\lim_{x\to1^{-}}=-5$.

4. 4

   Take the right-hand limit. For $x>1$, $x-1>0$ so $|x-1|=x-1$ and

   $\frac{(x-1)(2x+3)}{x-1}=2x+3\ \longrightarrow\ 2+3=5.$

   So $\displaystyle\lim_{x\to1^{+}}=5$. Note the sign pattern is the mirror image of the previous case: the side on which the numerator is positive has swapped.

5. 5

   Conclude. Because $-5\ne5$ the two-sided limit does not exist; the function has a jump of height $10$ at $x=1$. Contrast with $\lim_{x\to-2}\frac{2x^{2}+x-3}{|x-1|}$, where the denominator is nonzero and the limit is simply $\tfrac33=1$. The lesson: an absolute value only causes trouble when its argument vanishes at the point in question.