Limits, the old fashioned way

[Master_Nabo]

I understand that you will not do my homework for me, but I was hoping you would be able to push me in the right direction for this problem.  I am trying to solve the following limit:

lim                   x sin x
                      ----------
x -> 0             1 - cos x

This is an inderterminate form problem, so I attempted to find the Congugate Expression, but each time I come up with an answer, it is undefined.

I do not remember how to turn (1-cos x) into a (sin x) statement to cancel out some values.

I went around and did this problem on a calculator, and got the answer of 2, but I am wondering how to do this by hand, without using L'Hopitals or derivities.  I really could care less the answer to the problem, but I have spent the last hour trying to figure this out (its driving me nuts).

[Rule]

I understand that you will not do my homework for me, but I was hoping you would be able to push me in the right direction for this problem.  I am trying to solve the following limit:

lim                   x sin x
                      ----------
x -> 0             1 - cos x

This is an inderterminate form problem, so I attempted to find the Congugate Expression, but each time I come up with an answer, it is undefined.

I do not remember how to turn (1-cos x) into a (sin x) statement to cancel out some values.

I went around and did this problem on a calculator, and got the answer of 2, but I am wondering how to do this by hand, without using L'Hopitals or derivities.  I really could care less the answer to the problem, but I have spent the last hour trying to figure this out (its driving me nuts).

Why does it matter whether you use derivatives or not?  If you understand how L'Hopital's rule (or Taylor series) is derived, then each method is just as "concrete" as the other. 

x*sin(x)/(1-cos(x))  --> multiply top and bottom by 1+cos(x)
after some manipulation you'll get

lim x--> 0   x[1+cos(x)]/sin(x) =  lim x --> 0  x/sin(x) * lim x--> 0 [1+cos(x)]

lim x--> 0 [1+cos(x)] = 2.

Now we just need to find  lim x--> 0 x/sin(x).

I'm not sure if there is an easy way to reduce this without using derivatives...

For example, the taylor series of sin(x) = x - x^3/3! + x^5/5! + ...

so you've got

lim x --> 0  x / sin(x)  =  x/[x-x^3/3! +...]
=  lim x --> 0    1 / [1-x^2/3! + ... ]   = 1

and you're done.

Or you could just use L'Hopital's rule....

Here is my proof (there may be errors as this is mostly individual work):
The Mean Value Theorem:
If a function w is continuous, real valued, and differentiable on the real interval [a,b], there exists a c in [a,b] such that

w'(c) = [w(b)-w(a)] / [b-a].  Essentially the derivative at some point along the curve is equal to the slope of the secant line from a to b (this is a bit intuitive).


Apply the mean value theorem to
r(x) = f(x)[g(b)-g(a)] - g(a)[f(b)-f(a)]

You will find that
f'(c) / g'(c)  = [f(b)-f(a)] / [g(b)-g(a)]

Now imagine f(a) and g(a) --> 0.
f'(c) / g'(c) = f(b)/g(b) .   Now since c is in [a,b], if we let b-->a, then c must also approach a.
So, f'(b)/g'(b) = f(b)/g(b) so long as f(b) , g(b) --> 0 as b --> a.

This is also why L'Hopital's rule works if the numerator and denominator both approach something infinite (e.g. numerator could --> infinity, denominator could approach --> -infinity):

Assume w(b) / m(b) = infinite value / infinite value as b-->a

[1/w(b)]  / [1/m(b)] =  0/0 as b--> a  = d/db [ 1/w(b) ] /  d/db[1/m(b) ]
Using the chain rule,
m(b)/w(b)  = [-w'(b)/w(b)²] / [-(m'(b)/m(b)²]
= w'(b)*m(b)² / [w(b)²*m'(b)]
or,
w(b)/m(b) = w'(b)/m'(b)  as b--> a.

[Master_Nabo]

Thanks for the help, I understood most of it, but my algebraic skills are a bit rusty.  I understood that

Code Select Expand

x*sin(x)/(1-cos(x))  --> multiply top and bottom by 1+cos(x)
after some manipulation you'll get

lim x--> 0   x[1+cos(x)]/sin(x) =  lim x --> 0  x/sin(x) * lim x--> 0 [1+cos(x)]

But I didn't understand how you ended up with

Code Select Expand

im x--> 0   x[1+cos(x)]/sin(x)

I am just confused how you algebraically moved the (sin (x)) into the denominator, and put the (1+cos(x)) into the numerator.  I understand that you can move values between the numerator and denominator by changing the exponent, but I don't know which way you used.

After that step, I understood how you broke the function up, and got the answer.  The reason I did not want to bring in derivitives or L'Hopitals rule was to do it without them (primarily, this is the beginning of the semester, and the professor does Limits as his first unit, and Derivities as his second), and he assumes that we do not know of the other ways.  I just wanted to do the problem like the rest of the students who do not know of the other ways. 

[Rule]

x*sin(x) / [1-cos(x)] = x*sin(x) * [1+cos(x)] / [(1-cos(x))*(1+cos(x))]
= x*sin(x)*[1+cos(x)]/[1-cos²(x)] = x*sin(x)*[1+cos(x)]/[sin²(x)] = x*[1+cos(x)]/sin(x)

[Master_Nabo]

Thank you, you have made this so much clearer (and know I can finally end my day in peace).

[Joe[x86]]

Maybe I'm just stupid but I left more confused then I came in. ^_^

[rabbit]

Go take Calculus.