So I thought that whenever the denominator of a function was $0$ then the limit does not exist. Until I read the arithmetic rules for limits of functions that states if $f$ and $g$ are functions, and their limits are respectively $L$ and $M$. Then the limit of $f(x) \over {g(x)}$ as $x \rightarrow a$ is $L \over M$, if $M$ is not $0$. BUT if we assume $f(x) \over {g(x)}$ has a limit then the limit of $f(x)$ HAS to be $0$.
So I don't really understand this rule. I've always been taught (unless I'm just remembering wrong) that whenever the denominator is $0$, the function's limit does not exist. But now it does? How come $0 \over 0$ is defined all of a sudden? What's the difference between the limit of $2 \over {x-2}$ and $0 \over {x-2}$ as $x \rightarrow 2$?
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$\begingroup$Just because the denominator is approaching $0$ doesn't mean the limit does not exist. For example, consider the following three examples:$$\lim_{x \to 0}\frac{x}{x}=A, \quad \lim_{x \to 0}\frac{x^2}{x}=B, \quad \lim_{x \to 0}\frac{x}{x^2}=C.$$In all the three cases, the denominator is approaching $0$. However, $A=1, B=0$ and $C$ does not exist.
How the function behaves will depend on both numerator and denominator. In a manner of speaking, if they are both approaching $0$ then you may think of it as a race between the numerator and the denominator to see who reaches $0$ faster.
$\endgroup$ $\begingroup$Can limit exist in the form $\frac{0}{0}$?
The answer is Yes. You can find some simple examples in the other answer. And here is a more interesting one: $\lim_{x\to 0} \frac{\sin x}{x}=1$, while the denominator approaches $0$ as $x$ approaches $0$.
Can limit $\lim_{x\to a} \frac{f(x)}{g(x)}$ exist (in $\mathbb{R}$) if the denominator $g(x)$ approaches $0$ but $f(x)\to r\neq 0$ as $x\to a$?
No, because in this case the $\frac{f(x)}{g(x)}$ term can be infinity large(negatively or positively).
The idea of the definition is that it points out the ONLY possibility for the limit to exist.
$\endgroup$ $\begingroup$What we do is we algebraically determine what the expression to the right of the limit is first. Then we take a limit once we have an expression which we know how to handle.
$\lim_{x\to 2} \frac{0}{2-x} $ tells us what the expression $\frac{0}{2-x}$ approaches as $x$ approaches 2, not what it is at 2. So when we deal with this expression, we assume $x\not = 2$. Then 0 divided by any nonzero number is 0. So the limit is now $\lim_{x\to 2} 0=0$.
$\lim_{x\to 2} \frac{2}{2-x} $ tells us what the expression $\frac{2}{2-x}$ approaches as $x$ approaches 2. This one is trickier to understand fully. For a limit to exist, it must take either a definite value or it must approach the “same infinity” from both sides of the approached number. What I will show is that it cannot approach any particular value and it does not approach the same infinity.
Notice that when $x>2, \frac{2}{2-x}<0$ since the denominator is negative. Now notice that when $x<2, \frac{2}{2-x}>0$. So if both sides were to approach the same number, that number would have to be 0.
We also know that the expression cannot approach the same infinity since the values of $\frac{2}{2-x}$ have different sign on both sides of 2. So let’s proceed in showing it cannot approach 0.
Notice that if $0<x<2$, then $0>-x>-2$. So $2>2-x>0$ and $1/2<\frac{1}{2-x}<\frac{2}{2-x}$. This shows that no matter how close $x$ gets to 2 from the left (and to the right of 0), $\frac{2}{2-x}$ will always be at least 1/2. So in particular it will not approach 0. Thus, the limit does not exist.
$\endgroup$ $\begingroup$So what you're dealing with is called an indeterminate form. The indeterminate forms are strange to look at because there's no immediately clear answer to them when they come up. $0/0$ looks weird because a numerator of zero usually means the number zero, but a denominator of zero usually means infinity, so what's going on here? Well, when you get an expression in the form of $0/0$ you check to see whether the numerator or denominator goes to zero faster. How do you check that rate of change? You use the derivative (AKA the operation that can give you the rate of change of a function). If you call the numerator $f(x)$ and the denominator $g(x)$ then when you have $f(x)/g(x)$ in the form $0/0$ you take the derivative of both functions, $f'(x)/g'(x)$, and this gives the ratio of their rates of change.
This is a really useful trick called L'Hospital's rule. It's basic premise is that if you get an indeterminate form (0/0, infinity/infinity, infinity-infinity, infinity^0, 0^0, 1^infinity) and can get it to one of the ratio indeterminate forms (0/0, infinity/infinity) then you can apply the rule (lim $f(x)/g(x)$ = lim $f'(x)/g'(x)$). So it's a cool way to solve some more problems.
$\endgroup$ $\begingroup$If you need to calculate the following limit
$$\large{\lim_{x\to a}\frac{f(x)}{g(x)}}$$ and if you have that $\large{\lim_{x\to a}g(x)}=0$
If you can write $g(x)$ as a factor of $f(x)$, that is $f(x) = g(x)\cdot h(x)$ you will have
$$\large{\lim_{x\to a}\frac{f(x)}{g(x)}} = \lim_{x\to a}\frac{g(x)\cdot h(x)}{g(x)} = \lim_{x\to a}h(x)$$
And note that if $f(x) = g(x)\cdot h(x) \to 0$ since $g(x) \to 0$ And this say that if denominator tends to $0$, the limit exist if numerator tends to $0$ and a important condition is that $\lim_{x\to a}h(x)$ must also exist.
Consider the counterexample:
$\large{\lim_{x \to 0}\frac{f(x)}{g(x)}}$ with $g(x) = x$ and $f(x) = x\sin(\frac{1}{x}) +\sin(x)$ you can note the following things:
Denominator $g(x)$ tends to $0$
Numerator $f(x)$ tends to $0$
$g(x)$ can be expressed as a factor of $f(x)$ as $f(x) = \color{blue}{g(x)} \cdot \color{red}{h(x)} = \color{blue}{x} \cdot (\color{red}{\sin (\frac{1}{x}) + \frac{\sin (x)}{x}})$
but anyway, the limit does not exist.
What fails here?
Here fails the condition $\lim_{x\to a}h(x)$ does not exist.
In resume, when denominator tends to $0$, the limit exist $\iff$
Numerator tends to $0$ and the limit of $h(x)$ exist.
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