1. **State the problem:** Find the limit $$\lim_{x\to -4} \frac{x^2 - 15}{x^2 - 16}$$. 2. **Substitute the value directly:** Substitute $x = -4$ into the expression: $$\frac{(-4)^2 - 15}{(-4)^2 - 16} = \frac{16 - 15}{16 - 16} = \frac{1}{0}$$ Since the denominator is zero, direct substitution is undefined. 3. **Analyze the denominator:** The denominator is $x^2 - 16 = (x - 4)(x + 4)$. 4. **Check the behavior near $x = -4$:** - For values slightly less than $-4$, say $x = -4.1$, denominator $= (-4.1 - 4)(-4.1 + 4) = (-8.1)(-0.1) = 0.81 > 0$. - For values slightly greater than $-4$, say $x = -3.9$, denominator $= (-3.9 - 4)(-3.9 + 4) = (-7.9)(0.1) = -0.79 < 0$. 5. **Check numerator near $x = -4$:** Numerator $= x^2 - 15$. - At $x = -4$, numerator $= 16 - 15 = 1$ (positive). 6. **Determine the limit from left and right:** - As $x \to -4^-$, numerator $\to 1$ (positive), denominator $\to 0^+$, so fraction $\to +\infty$. - As $x \to -4^+$, numerator $\to 1$ (positive), denominator $\to 0^-$, so fraction $\to -\infty$. 7. **Conclusion:** Since the left and right limits are not equal, the limit does not exist. **Final answer:** The limit does not exist.