When two functions are equal, but not.

I haven't looked into it much, but this is something I've been aware of that I know I need to look into.

When I have a function $f(x)=\frac{x+1}{x+1}$, There is a discontinuity at $x=-1$, yet $\frac{x+1}{x+1}=1$ and has no discontinuity. It's like they're equal but not.

The qualities of the function are not preserved after the algebraic manipulation, so I can't strictly say that $\frac{x+1}{x+1}=1$.

This is an issue for me when understanding integrals. For instance, finding the definite integral of the quotient, if the discontinuity is within my limits, doesn't make sense. But after changing the quotient to a constant, it's possible: but I've found the area under a curve that wasn't complete. I've found a solution for an unanswerable, insensible question.

I hope I've made this clear. My question is, is this right? How do I come to terms with this?

Solution 1:

$x=-1$ is what's called a removable singularity of your function $f$: if you redefine your function at that point it becomes continuous.

As for integration, the definition of the Riemann integral does require the function to be defined everywhere in the interval in question. However, for a function which is undefined at some point you could consider an improper integral, which would converge to the correct value. When you get to Lebesgue integration, having the function undefined at some point is allowed, and won't make a difference.

Solution 2:

${x+1\over x+1}=1$ only when $x\ne -1$. Your function $f$ is not defined at $x=-1$.
The graph of $f$ has a "hole" in it:

Things can be made nice, though: You could redefine $f$ to take the value $1$ at $x=-1$. This would give you a new function $\tilde f$ that differs from $f$ at only one point. Also, $\tilde f$ is identically 1; and so, it can be integrated over any interval of finite length.

Strictly speaking the definite integral of $f$ over $[-2,1]$ is undefined; but, you should be able to convince your self that the area under the graph of $f$ over $[-2,1]$, say, is 3, since the area under the "hole" is 0. Better yet, imagine a rectangle of height 1 and very small width centered at $x=-1$. The area of that rectangle is very small; so the area under the graph of $f$ over $[-2,1]$ is 3 (after taking appropriate limits). This is what Paccio is doing in his answer.