hole on a continuous function

Question

Say f(x)= x+1.

Now $\lim \limits_{x \to 1}$ f(x), we can see easily that there is no hole in it or it is continuous everywhere. Since it is a straight line.

But when we write the same f(x) as,

f(x) = x+1 = (x+1)(x-1)/(x-1) = $(x^2-1)/(x-1)$.

$$\text{Or say g(x) }= (x^2-1)/(x-1)$$

i.e $\lim \limits_{x \to 1} g(x)=\lim \limits_{x \to 1}(x^2-1)/(x-1)$

In this site https://www.mathsisfun.com/calculus/limits.html it says g(x) has a hole at x=1. But isn't both f(x) and g(x) are same. Please explain. Thanks.

No, $g(x)$ isn’t the same as $f(x)$, because $g(x)$ isn’t defined at $x=1$, but it does have what is referred to as a “removable discontinuity”. — Joe, Sep 02 '19 at 11:37
$f$ and $g$ are the same for all $x\ne 1$. However, at $x=1$, the expression used to define $g$ would lead to the undefined "$\frac 00$", which we prevent by saying right away that $g$ is not defined at $x=1$ — Hagen von Eitzen, Sep 02 '19 at 11:38
Related: Why does factoring eliminate a hole in the limit?, Function continuity $(x^2 - 1)/( x - 1)$, Reasoning Behind Holes in Rational Functions — Martin R, Sep 02 '19 at 11:56

score 1 · Answer 1 · edited Sep 02 '19 at 13:02

1

A function isn't just an expression. It also has a domain (and usually a codomain, although that's less important for this discussion). Often when we write a function as an expression, we implicitly assume that the domain is as large as it could possibly be given the expression we have, rather than stating the domain explicitly. This is bad practice, but it is what it is.

With that in mind, the implied domain for $f$ is all of the real number line, while the implied domain for $g$ is the whole number line except $1$. When restricted to the domain of $g$, the two are indeed the same function, but the difference of domains makes them not the same.

edited Sep 02 '19 at 13:02

J. W. Tanner

60,406

answered Sep 02 '19 at 11:42

Arthur

199,419

Why they aren't the same.
Don't we write the equal('=') sign in between the expression i.e

x+1 = (x+1)(x-1)/(x-1).
– Rajesh Marndi Sep 02 '19 at 13:13
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@RajeshMarndi Yes, we do. But we need to be careful what that $=$ means. Or, more accurately, what the two things we compare actually are. As functions with maximal domains, no they aren't the same. On the more restricted domain, yes, they are equal (you might see something like $x\neq 1$ written next to such an equality to specify this). However, there are other views to take as well. One completely different view is to look at them not as functions, but as polynomials. And as pure, abstract polynomials, $\frac{x^2-1}{x-1} = x+1$ is as true as $\frac62 = 3$, with no caveats at all. – Arthur Sep 02 '19 at 13:20

hole on a continuous function

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