What to do when direct substitution tells us that limit does not exist but in actual it does exist?

Question

I am learning limits these days and I have reached at finding limits by direct substitution but there's one that doesn't making sense to me at all. Let me illustrate it with an example $f(x) = x/x$ .The limit ( as $x$ approaches $0$) in this case does exist but by direct substitution, it says limit does not exist. What to do when you come across this type of questions.

Answer 1

In the example you gave, as you're taking the limit when $\;x\to0\;$ , it is your right and duty to consider number "very close" to zero but different from zero. Under this understanding, we simply have $\;\cfrac xx=1\;$ for each number like this,and we can then easily deduce (according to its definition) the limit, which is 1.

In some other cases, as with $\;\cfrac{\sin x}x\;$ , taking close values of $\;x\;$ to zero gives values close to $\;1\;$ , yet the formal proof that $\;\lim\limits_{x\to0}\cfrac{\sin x}x=1\;$ is a little messier, though much prettier, and it is usually done by geometric arguments.

Resuming: every case must be considered separatedly, but sometimes, as in your example, you can do simple arithmetic manipulations and get your result pretty simply.

Answer 2

Use other methods!

I don't hope anybody has tried to tell you that substitution works when the expression isn't defined, because (as your example shows) it doesn't.

In this case, you just need a simple algebraic manipulation to transform the expression you had to find the limit of into something else that is defined at the point (and therefore it's different), where you can "substitute" (there are no $x$'s left so it's a very trivial substitution in this case).