We can apply l’Hôpital’s Rule to the indeterminate quotients $ \dfrac{0}{0} $ and $ \dfrac{\infty}{\infty} $, but why can’t we directly apply it to the indeterminate difference $ \infty - \infty $ or to the indeterminate product $ 0 \cdot \infty $?
Furthermore, why can’t we call $ \infty + \infty $ and $ \infty \cdot \infty $ indeterminate forms?
I’m new to calculus, so please clear up my concepts if you can. Thanks!
The phrase “indeterminate form” is used to mean a function that we can't compute the limit of by simply applying some general theorem.
One can easily show that, if $\lim_{x\to x_0}f(x)=a$ and $\lim_{x\to x_0}g(x)=b$, then $$ \lim_{x\to x_0}(f(x)+g(x))=a+b $$ when $a,b\in\mathbb{R}$. One can also extend this to the case when one or both $a$ and $b$ are infinite:
If $a=\infty$ and $b\in\mathbb{R}$, then $\lim_{x\to x_0}(f(x)+g(x))=\infty$
If $a=\infty$ and $b=\infty$, then $\lim_{x\to x_0}(f(x)+g(x))=\infty$
If $a=-\infty$ and $b\in\mathbb{R}$, then $\lim_{x\to x_0}(f(x)+g(x))=-\infty$
If $a=-\infty$ and $b=-\infty$, then $\lim_{x\to x_0}(f(x)+g(x))=-\infty$
(Note: $a$ and $b$ can be interchanged; $x_0$ can also be $\infty$ or $-\infty$.)
However, it's not possible to extend this to the case where $a=\infty$ and $b=-\infty$ (or conversely). We summarize this statement by saying that $\infty-\infty$ is an indeterminate form.
For instance,
if $f(x)=x$ and $g(x)=1-x$, then clearly $\lim_{x\to\infty}(f(x)+g(x))=1$
if $f(x)=x^2$ and $g(x)=x$, then $\lim_{x\to\infty}(f(x)+g(x))=\infty$
Other cases are possible.
There are similar criterions for functions of the form $f(x)/g(x)$; if the limits of the two functions exist, then we can easily say something about the limit of the quotient, except in the cases when
Therefore it's traditional to summarize this lack of general theorems in these cases by saying that $0/0$ and $\infty/\infty$ are indeterminate forms.
There's nothing mysterious: we just know that, in order to compute (or show the existence of) a limit that appears to be in one of the indeterminate forms, we have to do more work than simply calculate a quotient. For instance apply l'Hôpital's theorem, or cleverly rewrite the function, or using other methods such as Taylor expansion.
Also the case when $\lim_{x\to x_0}f(x)\ne0$ and $\lim_{x\to x_0}g(x)=0$ is somewhat delicate, but not really “indeterminate”: the limit of the quotient either doesn't exist or is infinite ($\infty$ or $-\infty$).
I'll try to shed some light on the somewhat obscure context in which "indeterminate forms" arise, even though this is far from may daily considerations and largely based on guessing about things that are not often clearly said (for instance not on Wikipedia). In fact my main encounter with indeterminate forms is them being brandished as ultimate (indeed unique) weapon in the battle waged by some to undefine $0^0=1$; more about that later.
The context is a short-cut to use continuity to avoid evaluating certain limits. By definition, $\lim_{x\to x_0}f(x)$ is insensitive to the value $f(x_0)$, or to whether it is defined at all. However, in case $f$ is continuous at$~x_0$, the limit will be equal to$~f(x_0)$, and computing that is in general easier than to apply the definition of the limit. The basic arithmetic operations $+$, $-$, $\times$, $/$ are continuous at all points where they are defined, as well as many functions like $\exp$, $\sin$, $\cos$, $\tan$, and if restricted to real numbers "$\sqrt{}$" as well as $\ln$ and inverse trigonometric functions like $\arctan$ (such functions can be extended to almost all complex numbers, but this involves choices that cannot be made throughout in a continuous way).
This takes care of many limits of an expression at a point where it is (continuously) defined, but of course the most interesting limits are those taken where the expression is not defined (with $x$ tending to infinity in some way or other, or to a point where the expression involves a non-defined case like division by$~0$).
Now the applicability of this method can be enlarged by extending both components of their domain$~\Bbb R^2$ of the operations, and their codomain, with additional points (at infinity), equipped with the appropriate topology, and then extending the operation by continuity where it can. For instance for addition one could add points $+\infty,-\infty$ to$~\Bbb R$ both in domain and codomain, with the usual neighbourhoods (which gives a Hausdorff topology, so limits when defined will be unique); one then defines addition with one infinite argument to return that argument, as well as $(+\infty)+(+\infty)=+\infty$, $~(-\infty)+(-\infty)=-\infty$. The resulting map defined in $\overline{\Bbb R}\setminus\{(+\infty,-\infty),(-\infty,+\infty)\}$ is continuous wherever it is defined. Similar extensions exist for subtraction and multiplication; in each case there are pairs of values where the operation must be left undefined if the map defined is to be continuous, since there are multiple values$~v$ such that every neighbourhood of these pairs contain other pairs where the operation gives$~v$ (indeed where this happens, it happens for _all_$~v$). For multiplication $(0,+\infty)$ is such a pair.
Division is slightly different, since it is already undefined for certain pairs in$~\Bbb R^2$, namely on the set $\Bbb R\times\{0\}$. In this case the operation can be continuously extended to most of this set by adding single point at infinity to the codomain and taking that as value for $a/0$ when $a\neq0$ (with two infinite points such an extension would not be possible). It turns out that the other arithmetic operations can also be extended using a single value$~\infty$ instead of $\pm\infty$, although one must then give up some defined cases like $(+\infty)+(+\infty)$, as $\infty+\infty$ cannot be continuously defined. On the other hand a function like $\exp$ can only be extended if $+\infty$ and $-\infty$ are distinguished. Some choices are therefore necessary as to how the topological spaces are extended; once this is done, the extension of operations/functions is limited only by the requirement of continuity.
Now if for some limit expression $\lim_{x\to x_0}f(x)$ the evaluation of $f(x_0)$ can be performed using (suitable chosen) extended operations, involving only cases where those operations are well defined, then the resulting must also give the limit, by continuity; if the resulting value should be $+\infty$ or $-\infty$, then this shows that the limit does not exist in$~\Bbb R$, and diverges in a specific way. In case the evaluation of $f(x_0)$ involves a case like $0/0$ where the extended operation is not (and cannot be continuously) defined, then this method fails, and some other method is needed to determine the limit (as a last resort, one could apply the definition of a limit). These are the infamous "indeterminate forms", they are just the points where the operations or functions used cannot be extended by continuity. What indeterminate forms precisely occur depends on which extensions are chosen. Some limits like $\lim_{x\to0}\exp(1/x)$ cannot be determined by this method, not because they hit an indeterminate form, but because no compatible extension of the codomains and domains is possible (in the example $1/0$ wants a single$~\infty$, but $\exp$ wants separate $\pm\infty$).
Continuity is the unique imperative of this method: all operations and functions involved must be continuous wherever they are defined. If some operation fails this requirement, the method becomes unreliable: in a point of discontinuity, the evaluation proceeds unstopped (because the operation is defined there) but the value produced might not be correct for the limit. For this reason one had to refrain from defining any value at the indeterminate forms, so evaluation will come to a halt when it finds one on the way. This brings me to the sad case of $0^0$, which is perfectly well defined as an empty product with value$~1$ (just like $0!$ is) or more generally as a case of the power$~0$, which gives the neutral element in any monoid. But if exponentiation is defined for positive real base and real (or complex) exponent by $x^y=\exp(y\ln x)$ then this becomes discontinuous at $(x,y)=(0,0)$ (only). So one would like to consider $0^0$ as an indeterminate form, but as I said, this requires that the value is undefined too, and this is why the fans of indeterminate forms which to undefine $0^0$. But $0^0=1$ is essential for many purposes, not in the least to allow $x^0$ to be replaced by $1$ whenever it occurs without having make an assumption $x\neq0$; undefining $0^0$ seems an unreasonable price to pay for validating something that is just a short-cut to avoid applying the definition of a limit. Instead I would propose that wherever a limit must be evaluated involving exponentiation where the exponent might be non-integer, such an expression $x^y$ be first replaced by $\exp(y\ln x)$, which can be safely evaluated with existing extended definitions (including $\ln0=-\infty$) without needing $0^0$ as indeterminate form; instead an indeterminate form $-\infty\times0$ might be hit inside the argument of $\exp$. In cases that are indeterminate in this way, the replacement of $x^y$ by $\exp(y\ln x)$ is usually the best way to proceed anyway.
Your first question is very simple: L'Hôpital's rule is a for things of the form $0/0$ and $\infty/\infty$. Therefore, if you have a problem not of these forms, you can't directly apply L'Hôpital's rule.
A better question would be to ask why L'Hôpital's rule does work for those two forms. IMO, the simplest explanation involves Taylor series and maybe Laurent series, although we can use differential approximation to get the flavor of what's going on.
We can write any differentiable function in the form
$$ f(x) = f(0) + x f'(x) + x r(x) $$
where $r(x)$ is a function (you can solve this equation for it) that satisfies
$$ \lim_{x \to 0} r(x) = 0 $$
Exercise: prove that the above is true. That is, solve for $r(x)$, and compute the limit.
In the case that $f(0) = 0$, this takes the special case
$$ f(x) = x f'(x) + x r(x) $$
Similarly, if we have another function $g(0) = 0$, we also have
$$ g(x) = x g'(x) + x s(x) $$
and so
$$ \frac{f(x)}{g(x)} = \frac{f'(x) + r(x)}{g'(x) + s(x)} $$
If $g'(0) \neq 0$ and $f'$ and $g'$ are continuous at $0$, then we can take the limit of the right hand side without trouble.
So L'Hôpital's rule can be thought of, in this case, as a way to cancel out the zero -- i.e. to cancel out one factor of $x$ -- from both the numerator and the denominator.
To extend this more generally, suppose we can write
$$ f(x) = x^n f_0(x) \qquad \qquad g(x) = x^n g_0(x) $$
where $g_0(0) \ neq 0$. Clearly, we have
$$ \lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{f_0(x)}{g_0(x)}$$
If we apply L'Hôpital's rule, we would get
$$\lim_{x \to 0} \frac{f(x)}{g(x)} = \lim_{x \to 0} \frac{n x^{n-1} f_0(x) + x^n f_0(x)}{n x^{n-1} g_0(x) + x^n g_0(x)} $$
So long as $f'_0(x)$ and $g'_0(x)$ do not diverge to $\pm \infty$ at $x=0$, it's easy to see we get the correct limit, by first canceling out an $x^{n-1}$ from everything.
This argument doesn't work in full generality, but it wasn't meant as a proof -- it was meant to convey that L'Hôpital's rule works, by cancelling things out in some subtle sense, which is why the two forms it applies two are both quotients, where you want to cancel something out (zeroes or a poles) to find the limit.
The answer is that their limits are not indeterminate. In fact, their limits are well-defined. Those limits may not be very useful computationally (for instance, $\infty-\infty$ and $\infty\times\infty$ both diverge), but they are meaningful limits nonetheless. On the other hand, $\frac00$ and $\frac\infty\infty$ do not have well-defined limits. Also, $0\times\infty$ will generally converge or diverge depending on which limit is "faster" (e.g. $\lim_{x\rightarrow \infty}x^2e^{-x}=0$ because exponents converge more rapidly than quadratics).
Recall that L'Hopital's Rule may not be applied for limits of the form $\frac1\infty$ or $\frac10$ either. This is because those limits are also well-defined (i.e. $0$ and $\infty$, respectively). L'Hopital's Rule only applies when the limit cannot be defined.
If a unique value be assigned to $f(a)$, then $f(x)$ is said to determinate at $x=a$, otherwise $f(x)$ is said to be indeterminate at $x=a$
For example , $$f(x)=\frac{x^2-9}{x-3}$$ $f(4)=7 ,$ which is unique Hence $f(x)$ is said to be determinate at $x=4$ But $$f(3)=\frac{9-9}{3-3}=\frac{0}{0}$$ Hence we can't assign a unique value to $f(3)$ $\frac{0}{0}=k ,$say $\implies 0=k.0$, which is true for any $k\in R$ .There is no unique value of $k$ which satisfy the above equation . Hence it is an indeterminate quantity.
