Evaluate the limit of (x^2 + 2x - 15)/(x^2 - 9) as x approaches 3.

Step-by-step solution

1. 1

   Try direct substitution first, and read the result. Putting $x=3$ into numerator and denominator gives

   $\frac{9+6-15}{9-9}=\frac{0}{0}$

   The $\tfrac00$ form is indeterminate, not undefined: it is a signal that numerator and denominator share the factor $(x-3)$, which can be removed.

2. 2

   Factor the numerator. You need two numbers multiplying to $-15$ and adding to $+2$; those are $+5$ and $-3$:

   $x^2+2x-15=(x+5)(x-3)$

3. 3

   Factor the denominator as a difference of squares.

   $x^2-9=(x-3)(x+3)$

   Both factorisations contain $(x-3)$, confirming the diagnosis from step 1.

4. 4

   Cancel the common factor. For $x\ne 3$ the fraction equals

   $\frac{(x+5)(x-3)}{(x-3)(x+3)}=\frac{x+5}{x+3}$

   Cancelling is valid here because a limit as $x\to 3$ never uses the value at $x=3$ — only values nearby, where $x-3\ne 0$.

5. 5

   Substitute into the reduced expression. The reduced function is continuous at $x=3$:

   $\frac{3+5}{3+3}=\frac{8}{6}=\frac{4}{3}$

   Numerically, at $x=2.9999$ the original quotient is $1.33334$ and at $x=3.0001$ it is $1.33333$, both closing on $4/3\approx 1.3333$.