Help with bit advanced math

[razi]

Hi guy, I have one task that I dont know how to solve.
Its about limit of a function. See picture.

First line is start, in second line i just extended my second fraction with (sinx)^2 and, as (sinx)^2/x^2 in limit to 0 goes to 1, in third line I omitted it. That is wrong however and gives me later bad result. My task is not to find good result but to know, why is this a bad procedure. I cant think of anything that we have learnt, so if there is someone who understands this here (I believe there should be) than please help. Thank you.

[xavier]

It's bad procedure because

(a)

(2 * sin^2 x) / (sin^2 x * x^2)

is not

(2 / sin^2) * (sin^2 x / x^2),

it's

2 * ( sin^2 x / (sin^2 x * x^2) )



(b) you are taking the limit as x --> 0 of the whole function, not just a part, so if you extract something from (or multiply something against) the second addend, you have to do the same to the first (the extracted/multiplied value needs to go outside the parens, in other words).



The first thing I see in the second line is that (sin^2 x / sin^2 x) = 1, so you're dropping out the wrong term anyway. The third line should have x^2 in the denominator, not sin^2 x. Note that since the second line is wrong anyway, you are going to need to multiply the additional multiplicand to the first addend as well, which will probably give you another sin/cos identity you can use.

[jjp]

The limit of ((sin x)^2) / (x^2) is not 1. (sin x) will never be greater than 1 or smaller than -1, therefor the same holds true for (sin x)^2. x^2 will approach zero as x approaches zero. From which follows: the limit of ((sin x)^2) / (x^2), when x approaches zero, is zero.

(2 * sin^2 x) / (sin^2 x * x^2)

is not

(2 / sin^2 x) * (sin^2 x / x^2),

it's

2 * ( sin^2 x / (sin^2 x * x^2) )

I think these two terms are identical. Or why shouldn't they be?

(b) you are taking the limit as x --> 0 of the whole function, not just a part, so if you extract something from (or multiply something against) the second addend, you have to do the same to the first (the extracted/multiplied value needs to go outside the parens, in other words).

He is multiplying the second addend by 1 (unless x = 0), which should be legal.

[xavier]

Actually, you get zero and infinity (because of the divide-by-zero singularity) fighting. "lim(x-->0) [ (sin^2 x) / x^2 ] == 1" is a standard mathematical identity (the proof escapes me atm, but it's one of those things we learned way back when).

[xavier]

(2 * sin^2 x) / (sin^2 x * x^2)

is not

(2 / sin^2 x) * (sin^2 x / x^2),

it's

2 * ( sin^2 x / (sin^2 x * x^2) )

I think these two terms are identical. Or why shouldn't they?

The trig identity used is only 1 as x approaches zero -- by itself it's relatively meaningless. So it's not like an "x/x" identity, and therefore it has to be applied to everything on the RHS, not just one term.

[jjp]

Actually, you get zero and infinity (because of the divide-by-zero singularity) fighting. "lim(x-->0) [ (sin^2 x) / x^2 ] == 1" is a standard mathematical identity (the proof escapes me atm, but it's one of those things we learned way back when).

Yes, you're right. I guess I'm too tired for math right now

[razi]

lim( f() - g() )= lim f() - lim g() and same for *

So lets look just at second fraction, putting 2 aside:

2lim( sin^2 x / (sin^2 x * x^2) )

sin^2 x / sin^2 x is always 1

Why is not (sin^2 x) / (sin^2 x * x^2) == (1/ sin^2 x) * (sin^2 x / x^2) , isnt that basic of multiplying fractions?
and then 2* ( lim(1/ sin^2 x) * lim(sin^2 x / x^2))

Sorry if its something easy that I dont see, we started derivation+limits just 2 weeks ago and I want to fully understand it.

[xavier]

So http://en.wikibooks.org/wiki/Calculus/Limits supports what you said on the limits -- it's been 15 years for me, so...

As for fractions, yes, sorry, multiplication works fine, it's addition in the denominator I was thinking of.

Going back to the second step tho -- that still smells fishy to me. By inspection, (2/x^2) != (2 / sin^2 x) -- you've changed the function is why you get different results. Maybe if you did the "limit of differences is the difference of limits" step and separated the two terms into two limits; I'm not sure you can remove that (sin^2 x / x^2) term until the limit is evaluated.

I'd recommend asking your teacher/prof on Monday -- if I go back to my workplace this weekend I'll bring back my calc textbook (yes, I still have all my textbooks and I recommend everyone keep theirs because you *never* know when you might need it later in life...but I digress) and see if it can provide a more definitive answer.

[razi]

Its definitely something with second step, including sin^2x is harmless but doing omitting (sinx)^2/x^2 is not, there must be some rules when is this step acceptable. So far I found out that fraction that I want to abbreviate must not be indefinite and so also, what is left= 2 / sin^2 must be finite which is I assume not, so that might be it. Im just not sure about any of this because Ive been doing a lot of such abbreviating lately and it never gave me bad results...