Evaluate the limit of (x^2 - 16)/(x + 4) as x approaches -4.

Step-by-step solution

1. 1

   Substitute to classify the limit. At $x=-4$,

   $\frac{(-4)^2-16}{-4+4}=\frac{0}{0}$

   The numerator vanishes at the same point as the denominator, so $(x+4)$ divides both and the discontinuity is removable rather than a genuine blow-up.

2. 2

   Factor the numerator as a difference of squares. Since $16=4^2$,

   $x^2-16=(x-4)(x+4)$

   This is the pattern $A^2-B^2=(A-B)(A+B)$ with $A=x$ and $B=4$.

3. 3

   Cancel $(x+4)$. For every $x\ne -4$,

   $\frac{x^2-16}{x+4}=\frac{(x-4)(x+4)}{x+4}=x-4$

   The original function and the line $y=x-4$ agree everywhere except at the single point $x=-4$, where the original is undefined — and a limit ignores that one point.

4. 4

   Evaluate the simplified linear function.

   $\lim_{x\to -4}(x-4)=-4-4=-8$

5. 5

   Read off what the graph looks like. The graph of the original expression is the straight line $y=x-4$ with a hole punched at $(-4,-8)$. The limit $-8$ is exactly the height of that hole, which is why the answer is finite even though the function itself has no value there.