극한의 성질과 계산

(a) When we evaluate the limit from the left, we only consider values of $x$ that are less than zero. For these values of $x$, our function is $x^2-3$. So, $\displaystyle \lim_{x \to 0^-} f(x)=\displaystyle \lim_{x \to 0^-} (x^2-3)=-3$.

(b) When we evaluate the limit from the right, we only consider values of $x$ that are greater than zero. For these values of $x$, our function is $x^2+3$. So, $\displaystyle \lim_{x \to 0^+} f(x)=\displaystyle \lim_{x \to 0^+} (x^2+3)=3$.

(c) Since the limits from the left and right do not agree, $\displaystyle \lim_{x \to 0} f(x)=$ DNE.

To further clarify the situation, the graph of $y=f(x)$ is sketched below:

(a) When we evaluate $\displaystyle \lim_{x \to -4^-}f(x)$, we only consider values of $x$ that are less than $-4$. For these values, $f(x)=x^3+8x^2+16x$. So, $$\lim_{x \to -4^-}f(x)=\lim_{x \to -4^-} (x^3+8x^2+16x)=(-4)^3+8(-4)^2+16(-4)=0$$ Note that, because $x^3+8x^2+16x$ is a polynomial, we can evaluate the limit by directly substituting in $x=-4$.

(b) When we evaluate $\displaystyle \lim_{x \to -4^+}f(x)$, we only consider values of $x$ that are greater than $-4$. For these values,

$$\begin{aligned} f(x)&=\frac{x^2+8x+16}{x^2+30x-4} \\ \text{So,} \\ \lim_{x \to -4^+}f(x)&=\lim_{x \to -4^+}\frac{x^2+8x+16}{x^2+30x-4} \\ \text{This is a rational function, and $x=-4$ is in its domain (we aren't doing anything suspect, like dividing by 0), so again we can directly substitute $x=-4$ to evaluate the limit:} \\ &=\frac{(-4)^2+8(-4)+16}{(-4)^2+30(-4)-4}=\frac{0}{-108}=0 \end{aligned}$$

(c) Since $\displaystyle \lim_{x\to -4^-}f(x)=\displaystyle \lim_{x \to -4^+}f(x)=0$, we conclude $\displaystyle \lim_{x \to -4}f(x)=0$.