# Topic:極限

## 概要

$\quad \lim _{x\to a}f(x)=L$

これは"関数f(x)においてxがaに限りなく近づくとき、その極限はLである"と読みます。 どうすれば関数f(x)でaにおいて極限が存在するかどうかを推測できるでしょうか。あるいは、どうすれば極限が、我々が後に取り上げる技術的なに定まるということを推測できるでしょうか。 これから我々はそれを直感的な見地から見ていきたいと思います。

まず我々が興味をもつのが関数$f(x)=x^{2}$ でxが2に近づく時の極限です。そんな興味深いものを、以下のような表記法を用いて書きあらわすことができます。

$\quad \lim _{x\to 2}x^{2}$

この極限を求めるのを試みるためのひとつの方法が、2に近い値を選んで2に近づいていくときに、f(x)をそれぞれ計算して何が起こるかを見てみるということです。 この操作は次のようにして行います。

 $x$ $f(x)=x^{2}$ 1.7 1.8 1.9 1.95 1.99 1.999 2.89 3.24 3.61 3.8025 3.9601 3.996

ここに2よりも小さい数字をとり、小さいほうから2に近づけていきます。あるいは2よりも大きな数字をとり、同様に大きいほうから2に近づけてみます。

 $x$ $f(x)=x^{2}$ 2.3 2.2 2.1 2.05 2.01 2.001 5.29 4.84 4.41 4.2025 4.0401 4.004

$\quad \lim _{x\to 2}x^{2}=4$

それでは別の例を見て見ましょう。我々が関数$f(x)={\frac {1}{x-2}}$ に興味を持っていると仮定して、xが2に近づいていくときの変化の様子を見てみましょう。極限表記では次のようになります。

$\quad \lim _{x\to 2}{\frac {1}{x-2}}$

その前に、xを小さい側、あるいは大きい側から2に近づけていったときの関数の値を計算してみましょう。次の表は小さい側から2に近づけていった表です。:

 $x$ $f(x)={\frac {1}{x-2}}$ 1.7 1.8 1.9 1.95 1.99 1.999 -3.333 -5 -10 -20 -100 -1000

つぎの表は大きい側から近づけていったものです。

 $x$ $f(x)={\frac {1}{x-2}}$ 2.3 2.2 2.1 2.05 2.01 2.001 3.333 5 10 20 100 1000

この場合、xを2に近づけていっても関数はいかなる値にも近づいていきませんね。このような場合、このことを「極限は存在しない」といいます。

これら2つの例はどちらも取るに足らないように思えるかもしれませんが、以下の関数について考えてみてください:

 $f(x)=\left\{{\begin{matrix}x^{2}&{\mbox{if }}x\neq 2\\{\mbox{undefined}}&{\mbox{if }}x=2\end{matrix}}\right.$ OR ${\begin{matrix}f(x)&=&{\frac {x^{2}(x-2)}{x-2}}\quad \end{matrix}}$ ## 厳密ではない極限の定義

 簡単な極限の定義Lを'xがcに限りなく近づくときのf(x)の極限といいます。 このとき、次のように書き表します $\lim _{x\to c}f(x)=L$ または $f(x)\to L\quad {\mbox{as}}\quad x\to c.$ ## 極限の法則

Now that we have defined, informally, what a limit is, we will list some rules that are useful for manipulating a limit. These will all be proven, or left as exercises, once we formally define the fundamental concept of the limit of a function.

{\begin{aligned}\lim _{x\to c}kf(x)&=&k\cdot \lim _{x\to c}f(x)&=&kL\\\lim _{x\to c}[f(x)+g(x)]&=&\lim _{x\to c}f(x)+\lim _{x\to c}g(x)&=&L+M\\\lim _{x\to c}[f(x)-g(x)]&=&\lim _{x\to c}f(x)-\lim _{x\to c}g(x)&=&L-M\\\lim _{x\to c}[f(x)\cdot g(x)]&=&\lim _{x\to c}f(x)\cdot \lim _{x\to c}g(x)&=&L\cdot M\\\lim _{x\to c}{f(x) \over g(x)}&=&{\lim _{x\to c}f(x) \over \lim _{x\to c}g(x)}&=&{L \over M}&{\hbox{as long as}}M\neq 0\end{aligned}}

これらの決まりはidentitiesとして知られており、; they are scalar multiplication, addition, subtraction, multiplication, and division of limits. (A scalar is a constant, and we say that when you multiply a function by a constant, you are performing scalar multiplication.)

Using these rules we can deduce some others. First, the constant rule states that if f(x) = b is constant for all x then the limit as x approaches c must be equal to b. In other words

$\lim _{x\to c}b=b$ .

Second, the identity rule states that if f(x) = x then the limit of f as x approaches c is equal to c. That is,

$\lim _{x\to c}x=c$ .

Third, using the rule for products many times we get that

$\lim _{x\to c}f(x)^{n}=\left(\lim _{x\to c}f(x)\right)^{n}$  for a positive integer n.

This is called the power rule.

### 例題

We need to simplify the problem, since we have no rules about this expression by itself. We know from the identity rule above that $\lim _{x\to 2}{x}=2$ . By the power rule, $\lim _{x\to 2}{x^{3}}=2^{3}=8$ . Lastly, by the scalar multiplication rule, we get $\lim _{x\to 2}{4x^{3}}=4\cdot 8=32$

To do this informally, we split up the expression, once again, into its components. As above,$\lim _{x\to 2}4x^{3}=32$ .

また、 $\lim _{x\to 2}5x=5\cdot \lim _{x\to 2}x=5\cdot 2=10$

であり、 $\lim _{x\to 2}7=7$ です。 これらを足し合わせて、

$\lim _{x\to 2}4x^{3}+5x+7=32+10+7=49.$

From the previous example the limit of the numerator is $\lim _{x\to 2}4x^{3}+5x+7=49.$  The limit of the denominator is

$\lim _{x\to 2}(x-4)(x+10)=\lim _{x\to 2}(x-4)\cdot \lim _{x\to 2}(x+10)=(2-4)\cdot (2+10)=-24.$

As the limit of the denominator is not equal to zero we can divide which gives

$\lim _{x\to 2}$  $(4x^{3}+5x+7) \over (x-4)(x+10)$  $=-{49 \over 24}$ .

We apply the same process here as we did in the previous set of examples;

$\lim _{x\to 4}$  $(x^{4}-16x+7) \over (4x-5)$  $=$  $\lim _{x\to 4}(x^{4}-16x+7) \over \lim _{x\to 4}(4x-5)$  $=$  $(\lim _{x\to 4}(x^{4})-\lim _{x\to 4}(16x)+\lim _{x\to 4}(7)) \over \lim _{x\to 4}(4x)-\lim _{x\to 4}(5)$

We can evaluate each of these; $\lim _{x\to 4}(x^{4})=256.$  $\lim _{x\to 4}(16x)=64.$  $\lim _{x\to 4}(7)=7.$  $\lim _{x\to 4}(4x)=16.$  $\lim _{x\to 4}(5)=5.$  $={199 \over 11}$

Evaluate the limit $\lim _{x\to 0}$  $1-\cos(x) \over x$ .

To evaluate this seemingly complex limit, be aware of your sine and cosine identities. We will also have to use two new facts. First, if f(x) is a trigonometric function (that is, one of sine, cosine, tangent, cotangent, secant or cosecant) and is defined at a, then $\lim _{x\to a}f(x)=f(a)$ . Second, $\lim _{x\to 0}$  $\sin(x) \over (x)$  $=1$ .

To evaluate the limit, recognize that $1-\cos(x)$  can be multiplied by $1+cos(x)$  to obtain $(1-cos^{2}(x))$  which, by our trig identities, is $sin^{2}(x)$ . So, multiply the top and bottom by $1+cos(x)$  (This is allowed because it is identical to multiplying by one). This is a standard trick for evaluating limits of fractions; multiply the numerator and the denominator by a carefully chosen expression which will make the expression simplify somehow. In this case, we should end up with:

$\lim _{x\to 0}$  $1-\cos(x) \over x$  $=$  $1-\cos(x) \over x$  $\cdot$  $1 \over 1$  $=$  $1-\cos(x) \over x$  $\cdot$  $1+\cos(x) \over 1+\cos(x)$  $=$  $[(1-\cos(x))\cdot 1]+[(1-\cos(x))\cdot \cos(x)] \over x\cdot (1+\cos(x))$  $=$  $1-\cos(x)+\cos(x)-\cos ^{2}(x) \over x\cdot (1+\cos(x))$  $=$  $1-\cos ^{2}(x) \over x\cdot (1+\cos(x))$  $=$  $\sin ^{2}(x) \over x\cdot (1+\cos(x))$  $=$  $\sin(x)\cdot \sin(x) \over x\cdot (1+\cos(x))$  $=$  $\sin(x) \over x$  $\cdot$  $\sin(x) \over (1+\cos(x))$

Your next step shall be to break this up into $\lim _{x\to 0}$ $\sin(x) \over (x)$ $\cdot$ $\lim _{x\to 0}$ $\sin(x) \over (1+\cos(x))$  by the limit rule of multiplication. By the fact mentioned above, $\lim _{x\to 0}$ $\sin(x) \over (x)$ $=1$ .

Next, $\lim _{x\to 0}$ $\sin(x) \over (1+\cos(x))$  $=$  $\lim _{x\to 0}\sin x \over \lim _{x\to 0}(1+\cos x)$  $=$  $0 \over 1+\cos 0$  $=0$ .

Thus, by multiplying these two results, we obtain 0.

We will now present an amazingly useful result, even though we cannot prove this yet. We can find the limit of any polynomial or rational function, as above, as long as that rational function is defined at c (so we are not dividing by zero). More precisely, c must be in the domain of the function.

 Limits of Polynomials and Rational functionsIf f is a polynomial or rational function that is defined at c then $\lim _{x\rightarrow c}f(x)=f(c)$ We already learned this for trigonometric functions, so we see that it is easy to find limits of polynomial, rational or trigonometric functions wherever they are defined. In fact, this is true even for combinations of these functions; thus, for example, $\lim _{x\to 1}(\sin(x^{2})+4\cos ^{3}(3x-1))=\sin 1^{2}+4\cos ^{3}(3(1)-1)$ .

### The Squeeze Theorem

The squeeze theorem is very important in calculus proofs, where it typically are used to confirm the limit of a function via comparison with with two other functions whose limits are known.

It is called the Squeeze Theorem because it refers to a function f whose values are squeezed between the values of two other function g and h. g and h both have the same limit L at a point a. As f are trapped between the values of the two functions that approach L, the values of f must also approach L.

A more mathematical definition is:

Suppose that $g(x)\leq f(x)\leq h(x)$  hold for all x in some open interval containing a, except possibly at $x=a$  itself. Suppose also that $\lim _{x\rightarrow a}g(x)=\lim _{x\rightarrow a}h(x)=L$ . Then $\lim _{x\rightarrow a}f(x)=L$  also. Similar statements hold for left and right limit.

## 極限を見つける

Now, we will discuss how, in practice, to find limits. First, if the function can be built out of rational, trigonometric, logarithmic and exponential functions, then if a number c is in the domain of the function, then the limit is simply the value of the function at c.

If c is not in the domain of the function, then in many cases (as with rational functions) the domain of the function includes all the points near c, but not c. An example would be if we wanted to find $\lim _{x\to 0}{\frac {x}{x}}$ , where the domain includes all numbers besides 0. In that case, we want to find a similar function, except with the hole filled in. The limit of this function at c will be the same, as can be seen from the definition of a limit. The function is the same as the previous except at a point c. The limit definition depends on f(x) only at the points where x is close to c but not equal to it. When $x\neq c$ , this condition doesn't hold, and so the limit at c does not depend on the value of the function at c. Therefore, the limit of the new function is the same as of the previous function. And since the domain of our new function includes c, we can now (assuming it's still built out of rational, trigonometric, logarithmic and exponential functions) just evaluate the function at c as before.

In our example, this is easy; canceling the xs gives 1, which equals x/x at all points except 0. Thus, we have $\lim _{x\to 0}{\frac {x}{x}}=\lim _{x\to 0}1=1$ . In general, when computing limits of rational functions, it's a good idea to look for common factors in the numerator and denominator.

Lastly, note that the limit might not exist at all. There are a number of ways in which this can occur:

"Gap": There is a gap (more than a point wide) in the function where the function is not defined. As an example, in

$f(x)={\sqrt {x^{2}-16}}$

f(x) does not have any limit when -4 ≤ x ≤ 4. There is no way to "approach" the middle of the graph. Note also that the function also has no limit at the endpoints of the two curves generated (at x = -4 and x = 4). For the limit to exist, the point must be approachable from both the left and the right. Note also that there is no limit at a totally isolated point on the graph.

"Jump": If the graph suddenly jumps to a different level, there is no limit. This is illustrated in the floor function (in which the output value is the greatest integer not greater than the input value).

Asymptote: In

$f(x)={1 \over x^{2}}$

the graph gets arbitrarily high as it approaches 0, so there is no finite limit. In this case we say the limit is infinite.

Infinite oscillation: These next two can be tricky to visualize. In this one, we mean that a graph continually rises above and below a horizontal line. In fact, it does this infinitely often as you approach a certain x-value. This often means that there is no limit, as the graph never approaches a particular value. However, if the height (and depth) of each oscillation diminishes as the graph approaches the x-value, so that the oscillations get arbitrarily smaller, then there might actually be a limit.

The use of oscillation naturally calls to mind the trigonometric functions. An example of a trigonometric function that does not have a limit as x approaches 0 is

$f(x)=\sin {1 \over x}.$

As x gets closer to 0 the function keeps oscillating between -1 and 1. It is also true that sin(1/x) oscillates an infinite number of times on the interval between 0 and any positive value of x. The sine function, sin(x), is equal to zero whenever x=kπ, where k is a positive integer. Between each value of k, sin(x) oscillates between 0 and -1 or 0 and 1. Hence, sin(1/x)=0 for every x=1/(kπ). In between consecutive pairs of these values, 1/(kπ} and 1/[(k+1)π], sin(1/x) oscillates from 0, to -1, to 1 and back to 0 and so on. We may also observe that there are an infinite number of such pairs, and they are all between 0 and 1/π. There are a finite number of such pairs between any positive value of x and 1/π which implies that there must be infinitely many between x and 0. From our reasoning we may conclude that as x approaches 0 from the right, the function sin(1/x) does not approach any value. We say that the limit as x approaches 0 from the right does not exist. (In contrast, in the case of a "jump", the limits from each side did exist. Since they weren't equal, though, the regular limit didn't exist.)

Incomplete graph: Let us consider two examples.

First, let f be the constant function f(q)=2 where we specify that q is only allowed to be a rational number. Let $q_{0}$  be some rational number. Then $\lim _{q\to q_{0}}f(q)=2$ , since, for q close to $q_{0}$ , f(q) is close to (in fact equals) 2.

Now let g be the similar-looking function defined on the entire real line, but we change the value of the function based on whether x is rational or not.

$g(x)=\left\{{\begin{matrix}2,&{\mbox{if }}x{\mbox{ is rational}}\\0,&{\mbox{if }}x{\mbox{ is irrational}}\end{matrix}}\right.$

Now g has a limit nowhere! For let x be a real number; we show that g can't have a limit at x. No matter how close we get to x, there will be rational numbers (where g will be 2) and irrational numbers (where g will be 0). Thus g has no limit at any real number!

## 漸近線を表すための極限表記の利用

Now consider the function

$g(x)={\frac {1}{x^{2}}}.$

What is the limit as x approaches zero? The value of g(0) does not exist; the value of g(0) is not defined.

$\qquad g(0)={\frac {1}{0^{2}}}$

Notice that we can make the function g(x) as large as we like, by choosing a small x, as long as x0. For example, to make g(x) equal to one trillion, we choose x to be 10-6. In this case, we say we can make g(x) arbitrarily large (as large as we like) by taking x to be sufficiently close to zero, but not equal to zero. And we express it algebraically as follows

$\lim _{x\to 0}g(x)=\lim _{x\to 0}{\frac {1}{x^{2}}}=\infty$

Note that the limit does not exist at x = 0.

Example: Finding Vertical and Horizontal Asymptotes

Use limits to find all vertical and horizontal asymptotes of the function $f(x)={\frac {1}{x+2}}\$

The graph of $f(x)={\frac {1}{x+2}}\$  has a vertical asymptote if f(x) goes to infinity. Therefore, we can determine any vertical asymptotes by setting the denominator equal to 0.

$x+2=0\$
$x+2-2=0-2\$
$x=-2\$

Notice that $f(-2)={\frac {1}{0}}$ . Therefore, f(x) has a vertical asymptote at x = -2.

The graph of $f(x)={\frac {1}{x+2}}\$  has a horizontal asymptote if f(x) converges on a value at infinity. We can find any horizontal asymptotes by finding $\lim _{x\to \infty }f(x)\ .$

$\lim _{x\to \infty }{\frac {1}{x+2}}={\frac {1}{\infty +2}}={\frac {1}{\infty }}=0$

Therefore f(x) has a horizontal asymptote at y = 0.

## 学習教材

To see the power of the limit, let's consider a moving car. Suppose we have a car whose position is linear with respect to time (that is, a graph plotting the position with respect to time will show a straight line). We want to find the velocity. This is easy to do from algebra, we just take the slope, and that's our velocity.

But unfortunately (or perhaps fortunately if you are a calculus teacher), things in the real world don't always travel in nice straight lines. Cars speed up, slow down, and generally behave in ways that make it difficult to calculate their velocities.

Now what we really want to do is to find the velocity at a given moment (the instantaneous velocity). The trouble is that in order to find the velocity we need two points, while at any given time, we only have one point. We can, of course, always find the average speed of the car, given two points in time, but we want to find the speed of the car at one precise moment.

Here is where the basic trick of differential calculus comes in. We take the average speed at two moments in time, and then make those two moments in time closer and closer together. We then see what the limit of the slope is as these two moments in time are closer and closer, and as those two moments get closer and closer, the slope comes out to be closer and closer to the slope at a single instant.