1

Limit law states that if a function is continuous,

\begin{equation*} \lim_{x \to c }f(x)=f(c) \end{equation*}

Why can't we use the above rule after fixing the removable discontinuity and state,

\begin{equation*} \lim_{x \to 2 } f(x)=f(2) = 6 \end{equation*} for

\begin{equation*} f(x)=\begin{cases} \frac{x^2+4x-12}{x^2-2x} \quad &\text{if} \, x \neq 2 \\ 6 \quad &\text{if} \, x = 2 \\ \end{cases} \end{equation*}

Geralt of Rivia
  • 13
  • Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. – Community Sep 19 '21 at 09:37
  • You can show (exercise) that $$\lim_{x\to 2}f(x)=4\neq 6.$$ – K.defaoite Sep 19 '21 at 09:44
  • I understand how to calculate the limit by looking at neighboring values of x at x=2. I'm trying to reconcile why the limit law does not apply in this case since we have made the function continuous by removing the discontinuity by defining an output value at x=2. – Geralt of Rivia Sep 19 '21 at 09:48
  • 1
    The point is not to define an arbitrary output value. Otherwise any function would be continuous. For your choice, the condition is violated as one of the other comments shows. Thus the function only gives rise to a cont. Function by filling in the correct value as dictated by the rule you quite (If this is possible). – Alexander Schmeding Sep 19 '21 at 09:57
  • @Spartan The function is defined at $x=2$ but it is not continuous at $x=2$. The limit law applies only to functions which are continuous. I think you are confusing a function being defined at a point and it being continuous there, which is not the same staement. – Maximilian Janisch Sep 19 '21 at 10:07
  • @Community Since when does this account post comments? :O – Maximilian Janisch Sep 19 '21 at 10:08
  • 1
    @MaximilianJanisch Thank you. That was the source of my confusion. – Geralt of Rivia Sep 19 '21 at 10:19

1 Answers1

1

First of all, I would hesitate to call the following a "limit law":

If $f$ is continuous at $c$, then $\lim_{x\to c}f(x)=f(c)$.

This is because the statement "$f$ is continuous at $c$" by definition means $\lim_{x\to c}f(x)=f(c)$. This "limit law" is not a theorem, but rather just what it means for $f$ to be continuous at $c$.

The function $x\mapsto\frac{x^2+4x-12}{x^2-2x}$ is a rational function (i.e. the quotient of two polynomial functions). With a little work, we can prove the following theorem:

Every rational function $f$ is continuous—that is, at every point $c$ where $f$ is defined, $\lim_{x\to c}f(x)=f(c)$.

The proof consists of three stages:

  • The "sum", "product", and "quotient" limit law. If $\lim_{x\to c}f(x)=l$, and $\lim_{x\to c}f(x)=m$, then:
    • $\lim_{x\to c}f(x)+g(x)=l+m$
    • $\lim_{x\to c}f(x)g(x)=lm$
    • $\lim_{x\to c}f(x)/g(x)=l/m$ (provided that $m\neq0$)
  • Using the above laws, prove that every polynomial function is continuous (see https://math.stackexchange.com/questions/732115/how-can-i-prove-that-a-polynomial-with-degree-n-is-continuous-everywhere-in)
  • Using the quotient law, prove that every rational function is continuous

Now, using the fact that rational functions are continuous, we see that if $c\neq0$ and $c\neq2$, then $$ \lim_{x\to c}\frac{x^2+4x-12}{x^2-2x}=\frac{c^2+4c-12}{c^2-2c} \, . $$ Unfortunately, this doesn't help us here as we are interested in the case $c=2$. However, by factorising, we see that for all $x$ such that $x\neq0$ and $x\neq2$, we have $$ \frac{x^2+4x-12}{x^2-2x}=\frac{(x-2)(x+6)}{x(x-2)}=\frac{x+6}{x} \, . $$ Let $g(x)=(x+6)/x$. Since there is a $\delta>0$ such that $f(x)=g(x)$ for all $x$ satisfying $0<|x-2|<\delta$, we must have $$ \lim_{x\to 2}f(x)=\lim_{x\to2}g(x) \, . $$ Since $g$ is a polynomial, it is continuous, i.e. $\lim_{x\to 2}g(x)=g(2)=4$. Now, suppose that we define the following function, as you did in your question (note that I am using $h$ instead of $f$): $$ h(x)=\begin{cases} \frac{x^2+4x-12}{x^2-2x} \quad &\text{if} \, x \neq 2\text{ and }x\neq0 \\ 6 \quad &\text{if} \, x = 2 \\ \end{cases} $$ Here, it is true that $\lim_{x\to 2}f(x)=\lim_{x\to 2}h(x)$ (again because there is a $\delta>0$ such that $f(x)=h(x)$ for all $x$ satisfying $0<|x-2|<\delta$). However, since $h$ is not a polynomial or a rational function, we do not know that $\lim_{x\to 2}h(x)=h(2)$. And, as it turns out, this is not the case.

However, since $\lim_{x\to 2}h(x)$ exists, it is possible to re-define $h$ so that it is continuous at $2$, here by setting $h(2)=4$. The fact that we can "patch" the removable discontinuity is why removable discontinuities are called "removable discontinuities" in the first place!

Joe
  • 19,636
  • Do you understand what "fixing the removable discontinuity" means?After "fixing the removable discontinuity" – user247327 Sep 19 '21 at 12:14
  • @user247327: Yes: it means re-defining $h$ at $2$ so that it is continuous at $2$. Here, we can set $h(2)=4$, and then $h:\mathbb R\setminus{0}\to\mathbb R$ is a continuous function. – Joe Sep 19 '21 at 12:17
  • 1
    @user247327: I'm not really sure why you asked that question. Is there something in my answer that you find lacking? – Joe Sep 19 '21 at 13:29
  • 1
    @Joe As a side remark, I am also usually very confused if I answer a question and get a comment as the one that you got above . – Maximilian Janisch Sep 19 '21 at 16:13