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For $\lim_{x\to c} f(x)$ you can just plug in the $c$ into $f(x)$ to get $f(c)$ and determine the limit. Why bother with all those formulas that break down the expression into smaller parts? It just seems like extra work to get to the same $f(c)$ result, even if the properties are mathematically sound. For the quotient formula, it does not seem to prevent a division by zero, either. I'm missing something big here. When do they actually come in handy?

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Your first statement is just false. That is not how you find limits. –  Chris Eagle Apr 30 '13 at 15:08
Per @Chris's comment: see this recent post:. Note also that there are conditions on when the properties are valid. –  amWhy Apr 30 '13 at 15:10
To your first sentence: What's $\lim_{x\to 0} \frac {\sin x}{x}$ ? –  AakashM Apr 30 '13 at 15:33
That you "can just plug in the $c$" is equivalent to $f$ being continuous at $c$ (or the definition of the latter in most calculus textbooks, if you wish). Not always true, of course. Take $f(x)=1$ for $x\neq 0$ and $f(0)=0$, for instance. Not even mentioning that $f$ needs not even be defined at $p$. –  1015 Apr 30 '13 at 15:41
@AakashM You would still be dividing by zero either way, right? The limit when using forula #4 in the OP would be 0/0 or undef. –  JackOfAll Apr 30 '13 at 18:27
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3 Answers

I think the whole point is to make things easier to solve. Imagine trying to solve this in your head in one go, as in considering the whole function $$\lim_{x \to \infty}[ (1+ {1 \over x})^{x} + {2x^2+4x+7 \over 4x^2+19x +7} +4]$$ impoossible yeah?

but by those laws this is just

$$\lim_{x \to \infty}[ (1+ {1 \over x})^{x}] + \lim_{x \to \infty}[{2x^2+4x+7 \over 4x^2+19x +7}] +\lim_{x \to \infty}[4]$$

which is much easier to see what the answer will be.

Here the middle term looks as though if we use the quotient rule it won't make sense as: $$\lim_{x \to \infty}[{2x^2+4x+7 \over 4x^2+19x +7}]= {\lim_{x \to \infty}2x^2+4x+7 \over \lim_{x \to \infty} 4x^2+19x +7} = {\infty \over \infty}$$

Even though this $\textit{does}$ make sense if you have seen l'hopitals rule we can work out the limit without it as we can use the first representation $\lim_{x \to \infty}[{2x^2+4x+7 \over 4x^2+19x +7}]$ and divide top and bottom by $x^2$.

My point is that they let you muck about with limits to put them into something 'nicer'.

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Plugging in $c$ into $f(x)$ to determine the limit $\lim_{x\rightarrow c}f(x)$ only works if $f$ is defined and continuous at $c$. From that point of view, the identities you listed say that if $f$ and $g$ are continous at a point $p$, then so is $f+g$, $f\cdot g$ and so on. This is not obvious, although it's not hard to prove either.

For the quotient formula, one of course needs to assume that $\lim_{x\rightarrow p} g(x) \neq 0$.

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Let $f: \mathbb{R}\to\mathbb{R}$ and let $a \in \mathbb{R}$, the equality

$$\lim_{x\to a}f(x)=f(a)$$

Is equivalent to saying that $f$ is continuous at $a$. In truth, you can expecte that a reasonable function behaves this way (that should be the natural behavior), so we separate functions with this behavior in their own class: the continuous functions. They have the property that if we move $a$ just a little bit, then we move $f(a)$ just a little bit (that's not mathematically precise, to make it precise you need the definition, but I'm just trying to show you what it means).

Now if you have a function like this one:

$$f(x)=\begin{cases}\phantom{-}x, & x>0, \\ \phantom{-}1, & x = 0 \\ -x, & x<0\end{cases}$$

Now, look that $\lim_{x\to 0}f(x) = 0$, however $f(0)=1$ which is not the limit. Look what the problem is: imagine you are approaching $0$ from the right, then function is going converging to zero. If you walk just a little bit closer to $0$, then $f(x)$ also get's just a little closer to zero. The problem is that when you get really close to zero, the function will simply jump to $1$, a little bit you walk on the direction of zero will not correspond anymore to walking a little bit in the direction of $0$. So, using the definition, there's some $\epsilon > 0$ such that for all $\delta >0$ is not true that rescriting $x$ to $(-\delta, \delta)$ will restrict $f(x)$ to $(-\epsilon, \epsilon)$.

So, continuous functions form a special class of functions due to this property of not "breaking" itself on the point. The properties on the other hand are useful when you have a function that you know the limit, and you want to calculate the limit of a function obtained by sum, multiplication, quotient and composition. For instance, if you know that the limit of $x$ is $a$ when $x$ goes to $a$, then it's direct from the properties that the limit of $x^2$ when $x$ goes to $a$ is $a^2$ because $x^2 = x x$, not that you didn't need to use the definition once again to show that because you have already shown it proving the properties in the general case.

I hope this helps you out. Good luck.

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