One last question on the concept of limits

I read through this post on the notion of limits Approaching to zero, but not equal to zero, then why do the points get overlapped? But there's one last question I have. [Trust me, it'll be my last ever question on limits!] It clearly says as you get closer and closer to a certain point you are eventually getting closer to the limiting point. I.e., as $\Delta x$ approches $0$, you get to the limit.

Let me give you an example which supports the above statement. Let's say you want to evaluate the limit $$ \lim_{x \to 2} \frac{x^2 - 4}{x-2} .$$ Sooner or later, you have to plug in $2$ and you get the answer $4$ and you say that the limit of that function as $x \to 2$ is $4$. But why should I plug in $2$ and not a number close to $2$? $x$ is certainly not equal to $2$, right?

Solution 1:

I would disagree that "we must plug in $2$". What is really going on is that you are breaking up the problem into simpler limits that can be done as if "by plugging in". Here are a few of the ideas that are going on "behind the scenes":

Theorem 1. Let $f$ and $g$ be functions defined on an open interval $I$ that contains $a$. If $f(x)=g(x)$ for all $x\in I$, except perhaps at $x=a$, then $$\lim_{x\to a}\;f(x) = \lim_{x\to a}\;g(x),$$ in the sense that either they both exist and are equal, or neither one exists.

The idea here is that the limit of a function depends only on what happens near the point, not at at the point. If you know the definition of a limit using $\epsilon$s and $\delta$s, this is more or less immediate, since we always exclude the case $x=a$ explicitly in that definition.

Theorem 2. Some easy limits:

1. $\displaystyle \lim_{x\to a}\; k = k$, where $k$ is a constant.
2. $\displaystyle \lim_{x\to a}\; x = a$.

Note that we are not "really" plugging in $a$ in these two limits; these are limits that can be computed using the definition of limit.

Theorem 3. Decomposing complicated limits into simpler ones: if $f$ and $g$ are functions, and $\lim\limits_{x\to a}\;f(x) = L$ and $\lim\limits_{x\to a}\;g(x)=M$, then:

1. $\lim\limits_{x\to a}\Bigl(f(x)+g(x)\Bigr) = L+M$.
2. $\lim\limits_{x\to a}\Bigl(f(x)-g(x)\Bigr) = L-M$.
3. $\lim\limits_{x\to a}\Bigl(\alpha f(x)\Bigr) = \alpha L$ for any constant $\alpha$.
4. $\lim\limits_{x\to a}\Bigl(f(x)g(x)\Bigr) = LM$.
5. If $M\neq 0$, then $\displaystyle \lim\limits_{x\to a}\;\frac{f(x)}{g(x)} = \frac{L}{M}$.

These are the things you use to determine $$\lim\limits_{x\to 2}\frac{x^2-4}{x-2}.$$ Let $f(x) = \frac{x^2-4}{x-2}$ and $g(x) = x+2$. Then $f(x)$ and $g(x)$ are different functions ($g(x)$ is defined everywhere, but $f(x)$ is only defined on $(-\infty,2)\cup(2,\infty)$ ), but at any $x\neq 2$, $f$ and $g$ have the same value. That means that, by Theorem 1, we must have $$\lim_{x\to 2}\;f(x) = \lim_{x\to 2}\;g(x).$$

Now, $\lim\limits_{x\to 2}\;g(x) = \lim\limits_{x\to 2}\bigl( x + 2\bigr)$. By Theorem 3, part 1, we know that $$\lim_{x\to 2}\;g(x) =\lim\limits_{x\to 2}\bigl(x + 2\bigr) = \left(\lim_{x\to 2}\; x\right) + \left(\lim_{x\to 2}\; 2\right),$$ provided both limits on the right exist. By Theorem 2 above, we know that $\lim\limits_{x\to 2} \;x = 2$ and $\lim\limits_{x\to 2}\;2 = 2$. So: $$\lim_{x\to 2}\;f(x) = \lim_{x\to 2}\;g(x) = \left(\lim_{x\to 2}\;x\right) + \left(\lim_{x\to 2}\;2\right) = 2 + 2= 4.$$ Each equality is justified by a theorem. Nowhere are we "plugging in", neither $2$ nor numbers close to $2$; we are evaluating limits using the definition and some results that afford us some short-cuts.

Where does that idea of "plugging in" come in? Because there is a class of functions for which limits can essentially be computed by "plugging in". These are the continuous functions.

Definition. Let $f(x)$ be a function, and let $a$ be a number. Then $f(x)$ is continuous at $a$ if and only if three things happen:

1. $f$ is defined at $a$;
2. $\lim\limits_{x\to a}\;f(x)$ exists; and
3. $\lim\limits_{x\to a}\;f(x) = f(a)$.

If $I$ is an interval, we say $f$ is "continuous on $I$" if it is continuous at every $a\in I$; we say $f$ is "continuous" if it is continuous at every point of its domain; we say $f$ is "continuous everywhere" if it is continuous at $a$ for every real number $a$.

So the functions that are continuous everywhere are precisely the functions whose limits can be computed by simply "plugging in" (evaluating), thanks to part 3 of the definition. Among the functions that are continuous everywhere are the constants and the polynomials (like $g(x) = x+2$ above). However, to prove that such functions are continuous and that you can evaluate their limits by simply plugging in, you must use the properties of limits mentioned above, which therefore cannot be established by simply "plugging in" (that would make the argument circular).

Solution 2:

To find a limit, we do not care what actually happens at $x = 2$. We only care about what the $y$-values of the function are doing as we get close to $2$. In other words, one way to think of the quantity $$ \lim_{x \to a} f(x) $$ is the following: How do the $y$-values of the function $f(x)$ behave when $x$ gets close to $a$? If the $y$-values get close to one particular $y$-value (call it $L$), then we say $\displaystyle\lim_{x \to a} f(x) = L$.

In your particular example, the function $\frac{x^2-4}{x-2}$ looks like the line $x+2$, except it has a hole in it:

In other words, we have that $\frac{x^2 - 4}{x-2} = x+2$, except when $x = 2$. The fact the functions are not equal everywhere does not bother us, since we are only concerned with the $y$-value of the function when $x$ is close to $2$. Therefore, when computing limits (and only when computing limits), we can think of $\frac{x^2 - 4}{x-2}$ and $x+2$ as the same. Now, it is easy to see that the $y$-values of the function get arbitrarily close to $4$ when $x$ approaches $2$, and therefore, we can write

$$ \lim_{x \to 2} \frac{x^2-4}{x-2} = 4. $$

Solution 3:

Ok. First note that $$\lim_{x \to 2} \frac{x^{2}-4}{x-2} = \frac{(x+2) \cdot (x-2)}{(x-2)} = \lim_{x \to 2 } (x+2) = 4$$

Now if you want to plug in a number close to $2$, then try $x= 2 \pm \delta$, where $\delta \to 0$ and see what answer you get. You should get the same answer.

Solution 4:

This is an application of The Squeeze Theorem. A version of the theorem says that if $f(x)=g(x)$ for all $x\neq a$ then $\lim_{x\to a}f(x)=\lim_{x\to a}g(x)$ provided that $\lim_{x\to a}g(x)$ exists. So, for all $x\neq 2$ we have

$$ \frac{x^2-4}{x-2}=x+2. $$

Thus

$$ \lim_{x\to 2}\frac{x^2-4}{x-2}=\lim_{x\to 2}(x+2)=4. $$

You are really plugging 2 into the function $x+2$ and using Squeeze Theorem.

Solution 5:

An important point to consider is that you cannot actually "plug in" $2$ and get $4$.

When $x = 2$, you will have a division by zero, since $x - 2 = 0$.

When calculating $\frac{x^2 - 4}{x - 2}$ for $x = 2$ using a computer, you may get $4$ due to rounding errors when you have set $x$ to a value very close to 2. This is due to the very fact that the function approaches $4$ as $x$ approaches $2$.