함수의 극한과 연속 전체 문제

$\displaystyle\lim_{x \rightarrow -2^-} f(x)=16$: 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.

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.

Any polynomial of degree one or higher will go to $\infty$ or $-\infty$ as $x$ goes to $\infty$. So, we need a polynomial of degree 0–that is, $f(x)$ is a constant. One possible answer is $f(x)=1$.

This will be the case for any polynomial of odd degree. For instance, $f(x)=x$.

Many answers are possible: also $f(x)=x^{15}-32x^2+9$ satisfies $\displaystyle\lim_{x \rightarrow \infty} f(x) = \infty$ and $\displaystyle\lim_{x \rightarrow -\infty} f(x)= -\infty$.

Many answers are possible; the tangent function behaves like this.

If we let $f$ be my height, the time of my birth is $a$, and now is $b$, then we know that $f(a) \leq 1 \leq f(b)$. It is reasonable to assume that my height is a continuous function. So by the IVT, there is some value $c$ between $a$ and $b$ where $f(c)=1$. That is, there is some time (we called it $c$) between my birth and today when I was exactly one meter tall.

Notice the IVT does not say precisely what day I was one meter tall; it only guarantees that such a day occurred between my birth and today.

One example is $f(x) = \left\{ \begin{array}{ll} 0&\text{when }0 \leq x \leq 1\\ 2&\text{when }1<x \leq 2 \end{array}\right.$. The IVT only guarantees $f(c)=1$ for some $c$ in $[0,2]$ when $f$ is continuous over $[0,2]$. If $f$ is not continuous, the IVT says nothing.

Yes. This is a straightforward application of IVT.

No. IVT says that $f(x)=0$ for some $x$ between $10$ and $20$, but it doesn't have to be exactly half way between.

No. IVT says nothing about functions that are not guaranteed to be continuous at the outset. It's quite easy to construct a function that is as described, but not continuous. For example, the function pictured below, whose equation format is somewhat less enlightening than its graph: $f(x)=\left\{\begin{array}{ll} -\frac{26}{5}x+65,&10 \leq x < 15\\ -\frac{26}{5}x+91,&15 \leq x \leq 20 \end{array}\right.$.

True. Since $f(t)$ is continuous at $t=5$, that means $\displaystyle \lim_{t \to 5} f(t)=f(5)$. For that to be true, $f(5)$ must exist — that is, 5 is in the domain of $f(x)$.

True. Using the definition of continuity, $\displaystyle \lim_{t \to 5} f(t)=f(5)=17$.

In general, false. If $f(t)$ is continuous at $t=5$, then $f(5)=17$; if $f(t)$ is discontinuous at $t=5$, then $f(5)$ either does not exist, or is a number other than 17.

An example of a function with $\displaystyle \lim_{t \to 5}f(t)=17 \neq f(5)$ is $f(t)=\left\{ \begin{array}{lcr} 17&,&t \neq 5\\ 0&,&t=5 \end{array} \right.$, shown below.

Since $f(x)$ and $g(x)$ are continuous at zero, and since $g^2(x)+1$ must be nonzero, then $h(x)$ is continuous at 0 as well. According to the definition of continuity, then $\displaystyle \lim_{x \to 0}h(x)$ exists and is equal to $h(0)=\frac{0f(0)}{g^2(0)+1}=0$.

Since the limit $\displaystyle \lim_{x \to 0}h(x)$ exists and is equal to zero, also the one-sided limit $\displaystyle \lim_{x \to 0^+}h(x)$ exists and is equal to zero.