# Why can we change a limit's function/expression and claim that the limits are identical?

Say you have limit as $x$ approaches $0$ of $x$. You could just write it as $\frac{1}{\frac{1}{x}}$ and then the expression would be undefined. So what are you really doing when you "rearrange" an expression or function so its limit "works", and doesn't have any division by zero? Why can we suddenly change one expression to another, when they are not exactly the same, and say the limit is the same? For example, the expression $x$ is not equal to the expression $\frac{1}{\frac{1}{x}}$ because the latter is not valid for $x=0.$

• Who says $\lim_{x\to 0}\frac{1}{1/x}$ is undefined? It is pretty clearly $0$. – Henning Makholm Jul 20 '16 at 18:40
• @HenningMakholm I never said the limit was undefined, but the expression itself is at $x=0$. Since the expressions $x$ and $\frac{1}{\frac{1}{x}}$ are not equal, why do we assume their limits be equal? That is my question. – Max Li Jul 20 '16 at 18:41
• The two expressions are equal – user99914 Jul 20 '16 at 18:42
• x @Max: Please edit the question to fix "then the limit would be undefined", which must be a typo if you never said the limit is undefined. – Henning Makholm Jul 20 '16 at 18:42
• @ArcticChar No they aren't, because $\frac{1}{\frac{1}{x}}$ does not have a defined value at $x=0$. – Max Li Jul 20 '16 at 18:43

When we take the limit for $x \rightarrow 0$ of a function $f$, we are actually interested to what happens to $f$ in a punctured neighborhood of $0$. That is, we don't actually care about the behaviour of the function when $x$ is exactly $0$. This means that $f(x)=x$ and $g(x)=\frac{1}{\frac{1}{x}}$ have the same limit for $x \rightarrow 0$, because they are equal in a punctured neighborhood of $0$.

Note that this can be seen from the definition itself; that is, $\lim_{x \rightarrow 0} f(x)=l$ means

$$\forall \ \epsilon > 0 \ \ \exists \ \delta > 0 : 0 < |x| < \delta \Rightarrow |f(x)-l| < \epsilon.$$

If you take another function $g$ and you assume $g=f$ in a punctured neighborhood of $0$, it follows that $$\lim_{x \rightarrow 0} f(x)=l \iff \lim_{x \rightarrow 0} g(x)=l\ .$$

Note that the limit of $\frac{1}{\frac 1x}$ as $x\to 0$ is not undefined. Of course you cannot plug in $x = 0$ in this expression. However, it is not how limit works. In the definition of limit you need only care about the value of the function near that point, ($0$ in this case). Since

$$x = \frac{1}{\frac 1x}$$

holds in a punctured neighborhood of $0$, so

$$\lim _{x \to 0} x = \lim _{x\to 0} \frac{1}{\frac 1x} = 0.$$

You will be less confused (I hope) if you realize that when we talk about

$$\lim_{x\to c}f(x),$$

the function $f$ does not need to be even defined at $c$. You only need that $f$ is defined on a punctured neighborhood of $c$.

• I meant that the expression is undefined, so it is not equal to the expression of $x$. Since the expressions are different, why can we assume the limits are the same? – Max Li Jul 20 '16 at 18:44
• @MaxLi: Because the expressions have the same value in some neighborhood of $0$ (namely the entire real line) except for at $0$ itself. – Henning Makholm Jul 20 '16 at 18:45

Remark: sort of a long comment.

What you are rediscovering are so called removable singularities: Note that the functions $$f(x)=x$$ and $$\tilde{f}(x)=\frac{1}{\frac{1}{x}},$$ are not the same, as they domain of definition differs: the first is defined for all real numbers, whereas the second is not defined at zero (as you noticed).

But one can adjust the second by defining $$\bar{f}(x):=\begin{cases}\tilde{f}(x)&, x\neq 0\\0 &, x=0\end{cases}.$$ As explained in the other answers, $\lim_{x\rightarrow 0}\bar{f}(x)=0,$ which implies that the function is continuous at $0$ (as was clerly expected).

Such an adjustment is called removal of singularity.

• I think you are missing the point. Having a limit is different from the function being continuous. And you still cannot answer why $$g(x):=\begin{cases}\tilde{f}(x)&, x\neq 0\\0 &, x=0\end{cases}$$ will have the same limit as $\tilde f$, as again these two are not the same function. – user99914 Jul 20 '16 at 18:54
• What I am trying to say, is that the operation he is doing is not mathematically correct, as the resulting function is not the same as the starting, but that this error can be easily corrected by taking the natural continuation, which exists as the limit is the same, independently of this transformation.. That the limit as $x$ tends to zero is the same, is trivial. – b00n heT Jul 20 '16 at 18:59

When we are computing limits like $x\rightarrow a$, the value of the function at $x=a$ is irrelevant.

This means that if we are evaluating $\lim_{x\rightarrow a} f(x)$, and we know of a second function $g(x)$ which is equal to $f(x)$ everywhere except at $x=a$, then we can make the substitution

$$\lim_{x\rightarrow a} f(x) = \lim_{x\rightarrow a}g(x).$$

The reasoning is that $f$ and $g$ agree everywhere except at $x=a$, and limits don't depend on the value of the function at the limit point, so therefore the two limit values must be identical.

This principle is what allows us to transform, for example,

$$\lim_{x\rightarrow 0}\frac{x^2 +2x}{x} \Rightarrow \lim_{x\rightarrow 0}(x+2) = 2$$

where here $f(x)\equiv \frac{x^2+2x}{x}$ and $g(x) \equiv x+2$ are two functions that have exactly the same values everywhere except at the irrelevant location $x=0$.