Evaluate the limit of (4 - x^2)/(2x^2 + x^3) as x approaches -2.

Step-by-step solution

1. 1

   Test substitution. At $x=-2$ the numerator is $4-4=0$ and the denominator is $8-8=0$, so the form is $\frac{0}{0}$ and a shared factor must be present.

2. 2

   Factor the numerator. As a difference of squares, $4-x^2=(2-x)(2+x)$. Note the order: writing it as $-(x-2)(x+2)$ is equivalent but easier to get a sign wrong with.

3. 3

   Factor the denominator. Pull out the common $x^2$: $2x^2+x^3=x^2(2+x)$.

4. 4

   Cancel the common factor. For $x\neq -2$ the quotient equals $\frac{(2-x)(2+x)}{x^2(2+x)}=\frac{2-x}{x^2}$, and the point $x=-2$ itself is irrelevant to the limit.

5. 5

   Substitute into the reduced form. $\frac{2-(-2)}{(-2)^2}=\frac{4}{4}=1$.

6. 6

   Interpret the result. The original function has a removable discontinuity at $x=-2$: filling in the value $1$ would make it continuous there.