The proofs of limit laws and derivative rules appear to tacitly assume that the limit exists in the first place

Solution 1:

You're correct that it doesn't really make sense to write $\lim\limits_{h\to 0}\frac{f(x+h)-f(x)}{h}$ unless we already know the limit exists, but it's really just a grammar issue. To be precise, you could first say that the difference quotient can be re-written $\frac{f(x+h)-f(x)}{h}=2x+h$, and then use the fact that $\lim\limits_{h\to 0}x=x$ and $\lim\limits_{h\to 0}h=0$ as well as the constant-multiple law and the sum law for limits.

Adding to the last sentence: most of the familiar properties of limits are written "backwards" like this. I.e., the "limit sum law" says $$\lim\limits_{x\to c}(f(x)+g(x))=\lim\limits_{x\to c}f(x)+\lim\limits_{x\to c}g(x)$$ as long as $\lim\limits_{x\to c}f(x)$ and $\lim\limits_{x\to c}g(x)$ exist. Of course, if they don't exist, then the equation we just wrote is meaningless, so really we should begin with that assertion.

In practice, one can usually be a bit casual here, if for no other reason than to save word count. In an intro analysis class, though, you would probably want to be as careful as you reasonably can.

Solution 2:

The other answers are perfectly fine; just a perspective that can save your day in situations in which the existence of the limit is actually a critical point.

The crucial definition is the one of limsup and liminf: these are always well defined, and all you have to know at the moment are the following two properties:

1. $\liminf_{x \to x_0} f(x) \le \limsup_{x\to x_0} f(x) $
2. The limit of $f$ exist if and only if $\liminf_{x \to x_0} f(x) = \limsup_{x\to x_0} f(x) $, and in this case the limit agree with this value.

Now imagine you do your computation twice: firstly, you compute the liminf; then you compute the limsup. In both computations, as soon as you arrive to something that actually has limit (like $2x+h$), because of property (2) you can forget about the inf/sup story and just compute the limit.

Since with some manipulations you arrive to something that actually has limit, both calculations will give the same result and, because of property (2) again, the limit exist and coincide with the value you just computed.

Now this is not really the thing you should do if you are doing introductory analysis and you don't know liminf and limsup: formal properties of these two are slightly different from the formal properties of lim, and you could end up with an error. But as long as you don't "touch" the limit, and you just make some manipulation inside theimit, the same argument will carry on: if you end up with a well defined result, it is the limit :)

Solution 3:

What we have here should really be interpreted as multiple statements:

(1.) If $ \lim_{h \to 0} \frac{2hx + h^2}{h} $ exists then $ \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h}$ exists and is equal to $\lim_{h \to 0} \frac{2hx + h^2}{h} $.

(2.) If $ \lim_{h \to 0} [2x + h] $ exists then $ \lim_{h \to 0} \frac{2hx + h^2}{h}$ exists and is equal to $\lim_{h \to 0} [2x + h]$.

(3.) If $ \lim_{h \to 0} 2x$ exists then $ \lim_{h \to 0} [2x + h]$ exists and is equal to $ \lim_{h \to 0} 2x$.

(4.) $ \lim_{h \to 0} 2x$ exists and is equal to $ 2x $.

Note that once we have (4.) the "if" (conditional) part of (3.) is satisfied and so on all the way up to (1.). You can see that assuming that the limit exists in statements 1 to 3 is not a problem because you haven't used used that assumption to prove that it actually does. That would be circular logic and no good.

Your log example is different to this in the way that you don't have a statement that takes the role of statement (4.) above, which would allow you to escape the conditional. You have only proven that $\log(0) = 0$ IF $\log(0)$ exists, not that $\log(0)$ exists! This in itself is not an incorrect conclusion.