XV

of the argument except the integers 0 , 1 , 2 , 3 , etc., and for these integral values it has the value 0 . Now let us think of its limit when x = 3 . We notice that in the definition of the limit the value of the function at a (in this case, a = 3 ) is excluded. But, excluding f ( 3 ) , the values of f ( x ) , when x lies within any interval which (i) contains 3 not as an end-point, and (ii) does not extend so far as 2 and 4 , are all equal to 1 ; and hence these values approximate to 1 within every standard of approximation. Hence 1 is the limit of f ( x ) at the

value $3$ of the argument $x$, but by definition $f(3)=0$.

This is an instance of a function which possesses both a value and a limit at the value $3$ of the argument, but the value is not equal to the limit. At the end of XI. the function $x^2$ was considered at the value $2$ of the argument. Its value at $2$ is $2^2$, i.e. $4$, and it was proved that its limit is also $4$. Thus here we have a function with a value and a limit which are equal.

Finally we come to the case which is essentially important for our purposes, namely, to a function which possesses a limit, but no defined value at a certain value of its argument. We need not go far to look for such a function, ${\displaystyle \frac{2x}{x}}$ will serve our purpose. Now in any mathematical book, we might find the equation, ${\displaystyle \frac{2x}{x}}=2$, written without hesitation or comment. But there is a difficulty in this; for when $x$ is zero, ${\displaystyle \frac{2x}{x}}={\displaystyle \frac{0}{0}}$; and ${\displaystyle \frac{0}{0}}$ has no defined meaning. Thus the value of the function ${\displaystyle \frac{2x}{x}}$ at $x=0$ has no defined

meaning. But for every other value of $x$, the value of the function ${\displaystyle \frac{2x}{x}}$ is $2$. Thus the limit of ${\displaystyle \frac{2x}{x}}$ at $x=0$ is $2$, and it has no value at $x=0$. Similarly the limit of ${\displaystyle \frac{x^2}{x}}$ at $x=a$ is $a$ whatever $a$ may be, so that the limit of ${\displaystyle \frac{x^2}{x}}$ at $x=0$ is $0$. But the value of ${\displaystyle \frac{x^2}{x}}$ at $x=0$ takes the form ${\displaystyle \frac{0}{0}}$, which has no defined meaning. Thus the function ${\displaystyle \frac{x^2}{x}}$ has a limit but no value at $0$.

We now come back to the problem from which we started this discussion on the nature of a limit. How are we going to define the rate of increase of the function $x^2$ at any value $x$ of its argument. Our answer is that this rate of increase is the limit of the function ${\displaystyle \frac{(x+h{)}^2-x^2}{h}}$ at the value zero for its argument $h$. (Note that $x$ is here a "constant.") Let us see how this answer works

in the light of our definition of a limit. We have $\frac{(x+h{)}^2-x^2}{h}=\frac{2hx+h^2}{h}=\frac{h(2x+h)}{h}.$

Now in finding the limit of ${\displaystyle \frac{h(2x+h)}{h}}$ at the value $0$ of the argument $h$, the value (if any) of the function at $h=0$ is excluded. But for all values of $h$, except $h=0$, we can divide through by $h$. Thus the limit of ${\displaystyle \frac{h(2x+h)}{h}}$ at $h=0$ is the same as that of $2x+h$ at $h=0$. Now, whatever standard of approximation $k$ we choose to take, by considering the interval from $-\frac{1}{2}k$ to $+\frac{1}{2}k$ we see that, for values of $h$ which fall within it, $2x+h$ differs from $2x$ by less than $\frac{1}{2}k$, that is by less than $k$. This is true for any standard $k$. Hence in the neighbourhood of the value $0$ for $h$, $2x+h$ approximates to $2x$ within every standard of approximation, and therefore $2x$ is the limit of $2x+h$ at $h=0$. Hence by what has been said above $2x$ is the limit of ${\displaystyle \frac{(x+h{)}^2-x^2}{h}}$ at the value $0$ for $h$. It follows, therefore, that $2x$ is what we have called the rate of increase of $x^2$ at the value $x$ of the argument. Thus this method conducts us to the same rate of increase