극한의 성질과 계산

$-1$

$10$

$-2$

$-3/2$

$-7/8$

$0$

$0$

$1$

$9$

$3$

$5/8$

$1$

$\pi/180$

The statement $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=10$ tells us that, as $x$ gets very close to 3, $f(x)$ is 10 times as large as $g(x)$. We notice that if $f(x)=10g(x)$, then $\dfrac{f(x)}{g(x)}=10$, so $\displaystyle\lim_{x \rightarrow} \dfrac{f(x)}{g(x)}=10$ wherever $f$ and $g$ exist. So it's enough to find a function $g(x)$ that has limit 0 at 3. Such a function is (for example) $g(x)=x-3$. So, we take $f(x)=10(x-3)$ and $g(x)=x-3$. It is easy now to check that $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$ and $\displaystyle\lim_{x \rightarrow 3} \dfrac{f(x)}{g(x)}=\displaystyle \lim_{x \to 3}\frac{10(x-3)}{x-3}=\displaystyle \lim_{x \to 3}10=10$.

One way to start this problem is to remember $\displaystyle \lim_{x \rightarrow 0} \dfrac{1}{x^2}=\infty$. (Using $\dfrac{1}{x^2}$ as opposed to $\dfrac{1}{x}$ is important, since $\displaystyle\lim_{x \rightarrow 0}\dfrac{1}{x}$ does not exist.) Then by “shifting" by three, we find $\displaystyle \lim_{x \rightarrow 3} \dfrac{1}{(x-3)^2}=\infty$. So it is enough to arrange that $\dfrac{f(x)}{g(x)}=\dfrac{1}{(x-3)^2}$. We can achieve this with $f(x)=x-3$ and $g(x)=(x-3)^3$, and maintain $\displaystyle\lim_{x \rightarrow 3}f(x)=\displaystyle\lim_{x \rightarrow 3}g(x)=0$. Again, this is one of many possible solutions.

(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 $\displaystyle \lim_{x \to -4}f(x)=0$.