Question

I've read about the squeeze theorem and how it proves that sinx/x = 1 as x approaches to 0.

However, lim sinx as x approaches to 0 = 0, so we basically have 0/x = 1 which is a bit offsetting at first sight.

Does the fact that we can cancel both x values have anything to do with it, or is it just a singular special case? (or am I totally in the wrong direction?)
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In this problem - lim sin x/(cosx +1) = 0 (as x approaches 0), and from what I get that is because lim sinx=0, and 0/2 = 0.

But how would I know that this too isn't a special case? Because there is nothing to cancel?

Answer 1

Both sinx and x approach 0 on either side. So the limit of sinx/x by inspection is 0/0

But 0/0 is an indeterminate limit form. It could be 1, it could be 0 it could be something else. Thus we need advanced techniques to find its value.

Answer 2

For the other case
you can substitute x = 0 into sin(x) and cos(x) +1 directly which gives you 0/(1+1) = 0 and is clearly finite.

If the limit is not finite from observation (i.e. by subbing in you get 0/0 or infinity/infinity) it is an indeterminate form.

Answer 3

Oh, so its because in the first both of them approach 0, and with the second problem the denominator doesn't?

Answer 4

The denominator can approach 0 as long as the numerator doesn't.

Answer 5

if they do the same thing we can't tell which is more significant contributor - the numerator or denominator.

Answer 6

Alright, so we basically know to try different things when it ends up being 0/0? Thanks a lot for helping out

Answer 7

one method you'll learn is L'Hopital's Rule for finding those types of limits. (However you are not allowed to use that rule to prove sinx/x =1 since L'Hopital's Rule depends on the derivative of sin x which is calculated as a result of the sinx/x limit result)

Answer 8

Alright. I will look into it when the time comes. Was just interested as to how we know to particularly do that to that problem and not others. Thanks