함수의 극한과 연속 전체 문제

No horizontal or vertical asymptotes.

Horizontal asymptote at $y=-1$; no vertical asymptotes

$\infty$

$-\infty$

$-\infty$

$\infty$

Since the tangent line to the curve at point $P$ passes through point $P$, the curve and the tangent line touch at point $P$. So, they must intersect at least once. By drawing various examples, we can see that different curves may touch their tangent lines exactly once, exactly twice, exactly three times, etc.

Many answers are possible; here is one.

As $x$ gets closer and closer to 3, $y$ gets closer and closer to 10: this shows $\displaystyle\lim_{x \rightarrow 3} f(x)=10$. Also, at 3 itself, the function takes the value 10; this shows $f(3)=10$.

Many answers are possible; here is one.

Note that, as $x$ gets closer and closer to 3 except at 3 itself, $y$ gets closer and closer to 10: this shows $\displaystyle\lim_{x \rightarrow 3} f(x)=10$. Then, when $x=3$, the function has value 0: this shows $f(3)=0$.

$\displaystyle\lim_{x \rightarrow -2^-} f(x)=16$: in order for the limit $\displaystyle\lim_{x \rightarrow 2} f(x)$ to exist and be equal to 16, both one sided limits must exist and be equal to 16.

Not enough information to say. If $\displaystyle\lim_{x \rightarrow -2^+} f(x)=16$, then $\displaystyle\lim_{x \rightarrow -2} f(x)=16$. If $\displaystyle\lim_{x \rightarrow -2^+} f(x)\neq 16$, then $\displaystyle\lim_{x \rightarrow -2} f(x)$ does not exist.

$\displaystyle\lim_{t \rightarrow 0} \sin t=0$: as $t$ approaches 0, $\sin t$ approaches 0 as well.

$\displaystyle\lim_{x \rightarrow 0^+} \log x = -\infty$: as $x$ approaches 0 from the right, $\log x$ is negative and increasingly large, growing without bound.

$\displaystyle\lim_{y \rightarrow 3} y^2=9$: as $y$ gets closer and closer to 3, $y^2$ gets closer and closer to $3^2$.

$\displaystyle\lim_{x \rightarrow 0^-} \dfrac{1}{x}=-\infty$: as $x$ gets closer and closer to 0 from the left, $\dfrac{1}{x}$ becomes a larger and larger negative number.

$\displaystyle\lim_{x \rightarrow 3} \dfrac{1}{10}=\dfrac{1}{10}$: no matter what $x$ is, $\dfrac{1}{10}$ is always $\dfrac{1}{10}$. In particular, as $x$ approaches 3, $\dfrac{1}{10}$ stays put at $\dfrac{1}{10}$.

When $x$ is very close to 3, $f(x)$ looks like the function $x^2$. So: $\displaystyle\lim_{x \rightarrow 3} f(x) = \displaystyle\lim_{x \rightarrow 3} x^2=9$

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: