Suppose there be a function, $$ f(x) = \frac {x^2-1}{x-1} $$

For $x=1$, the function becomes un defined. But, in Algebra we know it is allowed to Cancel Denominator and numerator by the common factor and this would result in the same expression which is equivalent to the first one. But,

$$\require{cancel} f(x) = \frac{(x+1)\cancel{(x-1)}}{\cancel{(x-1)}} =y$$ $$\Rightarrow y = x + 1 $$

But, here when I see that Graph of both the equation they look same however the first one is undefined for x = 1 which is not with the case of the second one. But, just canceling a common term in numerator and denominator or multiplying them, changes the whole function why?

So, rules of algebra don't work? or there is a problem with my understanding?

  • 1
    $\begingroup$ Could you add pictures of the graphs? You might be making some mistake, like actually plotting $x^2 - \frac1x - 1$ instead of $\frac{x^2-1}{x-1}$ (which is what happens if you write x^2-1/x-1 into any reasonable graphing tool), or any of several other possible mistakes, and we can't tell if you don't give us more to go by. $\endgroup$
    – Arthur
    Commented Sep 17, 2018 at 13:43
  • $\begingroup$ @Arthur Okay let me add them, nice idea, btw :) $\endgroup$
    – user427802
    Commented Sep 17, 2018 at 13:44
  • $\begingroup$ @AbhasKumarSinha What do you mean? Both are straight lines: wolframalpha.com/input/?i=(x%5E2-1)%2F(x-1) wolframalpha.com/input/?i=x%2B1 $\endgroup$ Commented Sep 17, 2018 at 13:45
  • 1
    $\begingroup$ @AbhasKumarSinha they are not equivalent because of domain. There are two different functions with the same graph in range $\mathbb{R}-\{1\}$ $\endgroup$ Commented Sep 17, 2018 at 13:53
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    $\begingroup$ @AbhasKumarSinha No you don't (followed the rules of algebra). You can't short fraction if there is possible to have $0$ in denominator. You can do it if and only if you are sure that there is no $0$ in denominator. For example: $\frac{(x+1)(sinx + 2)}{sinx + 2} = x + 1$ But not in this case. $\endgroup$ Commented Sep 17, 2018 at 14:00

4 Answers 4


The functions are different because they have different domains.

A function is defined by its domain, codomain, and graph or, alternatively, by its domain, codomain, and a rule that specifies how elements of the domain are mapped to elements in the codomain.

The implied domain of the function defined by $$f(x) = \frac{x^2 - 1}{x - 1}$$ is the largest set of real numbers that do not make the denominator equal to zero, which is $$\{x \in \mathbb{R} \mid x \neq 1\} = \mathbb{R} - \{1\}$$
Thus, we should write that $f$ is the function $f: \mathbb{R} - \{1\} \to \mathbb{R}$ defined by $$f(x) = \frac{x^2 - 1}{x - 1}$$ Its graph is the line $y = x + 1$ with a hole at the point $(1, 2)$ since $$\lim_{x \to 1} f(x) = \lim_{x \to 1} = \frac{x^2 - 1}{x + 1} = \lim_{x \to 1} (x + 1) = 1 + 1 = 2$$

Its graph is the punctured line shown below.


Notice that at every point in the domain of $f$, the denominator does not equal to zero. Thus, we may divide by $x - 1$ to obtain $$f(x) = \frac{x^2 - 1}{x - 1} = \frac{(x + 1)(x - 1)}{x - 1} = x + 1$$ for each $x \in \text{Dom}_f = \mathbb{R} - \{1\}$.

The function $g: \mathbb{R} \to \mathbb{R}$ defined by $g(x) = x + 1$ is defined for every real number. Its graph is just the line $y = x + 1$.


While the two functions agree on the intersection of their domains, they have different domains. Therefore, they are different functions.


We can prove that both of these functions have the same graph for domain $D=\mathbb{R}-\{1\}$:

let $a \in D$ be any number from domain. Then we have:

$f(a) = a + 1 = a + 1 \cdot \frac{a-1}{a-1} = \frac{a^{2}-1}{a-1} = f(a)$

So the graph is the same for $D=\mathbb{R}-\{1\}$: but in general these two functions are different because of Domain.

  • $\begingroup$ There are no two functions given. To speak about a function we have to know the image and domain. $\endgroup$
    – Cornman
    Commented Sep 17, 2018 at 13:55
  • $\begingroup$ @Cornman But you can compute the widest domain for some formula and what I meant was: these two the widest domain are different for those formulas. Now right? $\endgroup$ Commented Sep 17, 2018 at 13:58
  • $\begingroup$ In my opinion the widest domain which fits $f(x)=\frac{x^2-1}{x-1}$ is indeed $\mathbb{R}$, since the pole can be "fixed". It is $\lim_{x\to 1} f(x)=2$. $\endgroup$
    – Cornman
    Commented Sep 17, 2018 at 14:03
  • $\begingroup$ @Cornman No it isn't. Becouse there are two different limits for $1^{-}$ and $1^{+}$ $\endgroup$ Commented Sep 17, 2018 at 14:05
  • $\begingroup$ But you can put $f(x) = \frac{x^{2}-1}{x-1}$ for $x \neq 1$ and $f(x) = 2$ for $x = 1$. Of course you can do it but the formula will be different. $\endgroup$ Commented Sep 17, 2018 at 14:06

The problem is, that $f(x)=\frac{x^2-1}{x-1}$ is not a function, for the simple reason, that a function has to be defined with a domain and image like:


One might consider $\mathbb{R}\setminus\{1\}$ as the domain of $f$, since for $x=1$ we get $1-1=0$ in the denominator, which leads to an illegal operation (dividing by zero). So this is a pole.

Or is it?

It is completly fine to define $f:\mathbb{R}\to\mathbb{R}$, since the pole is "liftable". A fake pole, if you please.


The only difference is that $\frac{(x^2-1)}{(x-1)} $ is not defined at x=1, although the limit as x tends to 1 exists and is equal to 2. Rest everything is same.


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