Evaluate the limit of (2x^2 + x - 3)/|x - 1| as x approaches -2.

Step-by-step solution

1. 1

   Check the denominator before doing anything clever. Absolute-value denominators usually signal a one-sided analysis, but only when the expression inside can vanish at the limit point. Here

   $|x-1|\ \text{is zero only at}\ x=1,$

   and the limit point is $x=-2$, comfortably away from it. So there is no indeterminate form to resolve.

2. 2

   Resolve the absolute value on a neighbourhood of $x=-2$. For $x$ near $-2$ we have $x-1<0$ (at $x=-2$ it equals $-3$), so on that whole neighbourhood

   $|x-1|=-(x-1)=1-x.$

   Rewriting removes the absolute value entirely and shows the function is a ratio of two polynomials near $x=-2$ — hence continuous there.

3. 3

   Confirm the form is not 0/0. Evaluating numerator and denominator separately at $x=-2$:

   $2(-2)^{2}+(-2)-3=8-2-3=3,\qquad |-2-1|=|-3|=3.$

   Both are nonzero, so no factoring, rationalising, or L'Hopital step is needed — those tools exist for indeterminate forms only.

4. 4

   Apply direct substitution. Since the function is continuous at $x=-2$, the limit is just the function value:

   $\lim_{x\to-2}\frac{2x^{2}+x-3}{|x-1|}=\frac{3}{3}=1.$

5. 5

   Confirm the two one-sided limits agree. Approaching from the left ($x=-2.001$): $\frac{2(4.004)+(-2.001)-3}{3.001}=\frac{3.007}{3.001}\approx 1.002$. From the right ($x=-1.999$): $\frac{2.993}{2.999}\approx 0.9980$. Both tend to $1$ as the step shrinks, so the two-sided limit exists and equals $1$ ✓. Contrast this with $\lim_{x\to1}\frac{2x^{2}+x-3}{|x-1|}$, where the absolute value really does force different left and right values.