165. If the dividend and divisor are simultaneously equal to zero, the quotient, according to the above definition of division (see the previous question), is equal to any number $m$, because $m \cdot 0 = 0$. Despite this, it sometimes happens that when the dividend and divisor simultaneously become equal to zero, the quotient still has a completely definite value. For example, for $x \rightarrow 0$, the expression $\frac{4 x}{x}$, although both the numerator and the denominator become zeros, is equal to 4. Provide more interesting examples and explain the precise meaning of the word "become."
165. Let the algebraic fraction be $\frac{x^{2}-3 x+2}{x^{2}-4 x+3}$. For $x=1$, both the numerator and the denominator become zero, and the expression takes the indeterminate form $\frac{0}{0}$, i.e., it has no definite value. Let's try to circumvent this difficulty in the following way: the numerator and the denominator can be factored into linear factors, so we can write: $\frac{x^{2}-3 x+2}{x^{2}-4 x+3}=$ $\frac{(x-1)(x-2)}{(x-1)(x-3)}$ or, by canceling $x-1,=\frac{x-2}{x-3}$. Let's substitute $x=1$ into this last expression and obtain the value $\frac{1}{2}$. Thus, the given algebraic fraction has a definite value for $x=1$, namely $\frac{1}{2}$. However, this method of operation has a significant flaw. We canceled the numerator and the denominator by $x-1$, but this is only permissible when it is known in advance that $x-1$ is not zero (division by zero is not allowed!), i.e., that $x \neq 1$, whereas we know the opposite: we are precisely looking for the value of the given expression for $x$. Therefore, we had no right to cancel by $x-1$. Let's now approach the question differently. For $x \neq 1$, we have unconditionally: $\quad \frac{x^{2}-3 x+2}{x^{2}-4 x}=\frac{x-2}{x-3} . \quad$ If $\quad x \quad$ approaches one infinitely (remaining always different from one), then the value of the expression
$\frac{x^{2}-3 x+2}{x^{2}-4 x+3}=\frac{x-2}{x-3}$ tends to the value $\frac{1-2}{1-3}=\frac{1}{2}$. Thus: although for $x=1$ the expression $\frac{x^{2}-3 x+2}{x^{2}-4 x+3}=\frac{0}{0}$, i.e., is meaningless, nothing prevents us from agreeing to call the "value of the fraction $\frac{x^{2}-3 x+2}{x^{2}-4 x+3}$ at $x=1$ " the limiting value to which this fraction tends when $x$ approaches one infinitely (becomes one).