Question

Okay, so I have the limit as x approaches 0 for the function (x+sinx)/x. I split it into two separate limits, both with x approaching 0, and the functions are x/x and sinx/x. I know if I plug in 0, x/x will give the indeterminate form. But since it says x/x, can I say that that limit will be 1? I know the answer is two, but that part bothers me.

Answer 1

\[\frac{x}{x}=1\] always (unless of course \(x=0\) but since you are taking a limit, you know \(x\neq 0\))

Answer 2

so yes, it is 1
what does \(\frac{x}{x}\) really mean? it means \(1\) except if \(x=0\) it is undefined

Answer 3

OH. Thank you for wording it that way. This whole unit, I never thought of it that way. When I took the limit as x approaches 0, I thought of it as x=0. I didn't really realize that it's just the path to x=0, not actually x=0.

Answer 4

1) Never, EVER "plug in" anything. Sometimes, a substitution is appropriate. It is appropriate with a limit only after you have demonstrated continuity.

2) It does not require any sort of substitution to observe an indeterminate form. "Indeterminate Form" is a concept, not a number or a value. You cannot build one or create one. You can only observe or define such a thing. 0/0 doesn't mean anything in the numerical world. In this world, a limit can by of TYPE 0/0, making it an indeterminate form.

3) Is there some reason to split it up? It already has the 0/0 indeterminate form.

My views. I welcome others'.

Answer 5

it is definitely not zero
that is the entire reason for all this limit business
you want to say "what it gets close to" without actually being that number

Answer 6

historically derivatives came first, limits second
people already knew how to take derivatives, but the language was imprecise, because without the notion of a limit, it didn't really make sense to say "it gets smaller and smaller but is never actually that number, but we pretend it is anyways"

Answer 7

i have to disagree somewhat with @tkhunny
you do "plug in 0" to see that
\[\frac{\sin(x)}{x}\] for example is of the form \(\frac{0}{0}\)
without plugging it in, how would you know?

Answer 8

in fact, i would always plug in the number first
if you get a number out, that is almost always the limit
most any function that you know is continuous on its domain, unless it is defined say as a piecewise function

if i wanted
\[\lim_{x\to \pi}\frac{\cos(x)}{x}\] i would immediately plug in \(\pi\)

Answer 9

and even if i wanted
\[\lim_{x\to \pi}\frac{\cos(x)+1}{x-\pi}\] i would still plug it in
once i got \(\frac{0}{0}\) i would know that it was of that form

Answer 10

Well, this is why we leave it open to discussion. :-)

1) "Plug in" doesn't exist. It's called "substitution".
2) We consider the limits of numerator and denominator separately. No substitution required.