Calculus halp?

[Kaelia]

I'm having sort of a hard time taking care of a calculus problem I was given. It's not really a very good subject to begin with for me (god only knows how I managed a B in a fast summer course of pre-cal), but it doesn't help that I have no way to check my answer and my professor refused to help me when I went in to ask him to check my work. Anyway, the problem is:

lim (cos pi x+1)/x-1 Parentheses just in place to show that it's all over x-1.
x -> 1

I tried substitution. As x -> 1, w -> 0 and changed everything in terms of w. w = x-1 and x = w+1.

lim (cos pi (w+1)+1)/w
w -> 0

Distributed cos pi into the (w+1). (cos pi w + cos pi + 1)/w. (Had a problem here not knowing whether or not the "cos pi x" was a single entity or the "cos pi" and "x" were separate. Assumed they were separate, but no clue, and professor wouldn't point me in the right direction.)

Since cos pi is -1 I just canceled the cos pi and the +1 in the numerator. (Cos pi w)/w

Canceled the w in both the numerator and the denominator and was left with cos pi which is -1. (Had a problem here too because I wasn't sure if I could cancel the "w"s seeing as they represent 0's and 0/0 is an indeterminate quantity rather than being 1. I went with it anyway because at this point I was lost and anything that gave me an answer was good enough until I could talk to somebody who is actually willing to help.

He gave us another problem that not even another calculus teacher could solve, but luckily he's going to excuse us from that one for obvious reasons. Any help with this one however?

[Demosthenes11]

gotta clarify what you mean by cos pi x+1

[Kaelia]

gotta clarify what you mean by cos pi x+1

"cosπx+1" is exactly how the numerator was written in the problem he gave us. Not sure if I'm meant to see it as cosπ(x+1) or (cosπx)+1. I'm just not sure if he meant the first part as cosine of π or cosine of πx. I assumed just cosine of π. Sorry, I hope that's what you meant by clarifying. He assumed we'd understand it exactly as it was written.

Edit: Bah, pi symbol looks weird in this font. Sorry. π = pi

[The_OG_Nelta]

Aurik Signal~

[Bobbity]

As it is written, I would assume the numerator is cos(pi*x)+1 (which also makes the problem a bit simpler). Note that as x goes to 1, the denominator goes to 0 and the numerator goes to:


cos(pi*1)+1 = cos(pi)+1 = (-1)+1=0


Thus, we have an indeterminate form (0/0) and can use l'Hopital's rule. Namely, take the derivative of numerator and denominator and then take the limit. This gives:


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


where I used d/dx( cos(pi*x)+1 ) = -pi*sin(pi*x) and d/dx(x-1)=1. I hope this helps.

[Demosthenes11]

lol i seriously doubt first year calc would be at l'Hopital right now, gonna work on it in a sec

[JasoNintendo]

For the record, I did l'hopitals rule in the first year, so it could be that. Only real way to know exactly what is meant to be done is a clarification of the question, but that's tough if it's set up like that.

[Jefe]

I'm going to assume the actual problem is: (cos (pi*x) + 1) /(x-1) as the lim of x->1. If you plug in the values you'll notice the problem it's indeterminate type (0/0). Therefore, you must use L'Hospital's rule: f'(x)/g'(x), that means find the derivative of the numerator and denominator.

Thus Dx(cos(pi*x)+1)/(x-1) = [(-Sin(pi*x)(pi) + 0)/(1)], plug in the value for x, you get: zeroth.

[Kohan]

Originally, I read this as "Cactus Help." I'm disappointed that's not what this thread is about.

[Kaelia]

We haven't learned l'Hopital's rule yet. I don't know if he thinks we learned it in pre-calculus or not. I'm also not sure what it means to find the derivative; haven't heard of that yet either. Can anyone explain to me basically how to do that so that it's actually plausible for me to use that method as my solution? Thank you guys very much for the responses so far.

[Demosthenes11]

lol don't worry about l'Hopital. You will learn that toward the end of the semester / next semester. You'll be doing derivatives in about a week as well, so don't worry about that either. I played briefly with this and couldnt really get it, but I haven't done elementary methods of limits in a long time.

Don't hate the class, gotta love math!

[Galois]

First, let us show that the limit as x approaches 1 of sin(pi*x)/(x-1) is -pi. This will be done by using a substitution.

Let x=(pi-t)/pi. This means that t=pi-pi*x. So, when x=1, t=0.

We now have:

lim_t->0 sin[pi(pi-t)/pi]/[(pi-t-pi)/pi]

=lim_t->0 sin(pi-t)/(-t/pi)

=-pi*lim_t->0 sin(pi-t)/t

=-pi*lim_t->0 sin(t)/t (since sin(pi-t)=sin(t))

=-pi*1=-pi.


Now that we know this fact, we can evaluate the limit.

lim_x->1 [cos(pi*x)+1]/[x-1]

=lim_x->1 [(cos(pi*x)+1)(1-cos(pi*x))]/[(x-1)(1-cos(pi*x))] (multiply the numerator and denominator by (1-cos(pi*x))

=lim_x->1 (1-(cos(pi*x))^2)/[(x-1)(1-cos(pi*x))] (difference of squares)

=lim_x->1 (sin(pi*x))^2/[(x-1)(1-cos(pi*x))] ( since (sin k)^2 = 1-(cos k)^2)

=lim_x->1 sin(pi*x)/(x-1) * lim_x->1 sin(pi*x)/(1-cos(pi*x))

=(-pi) * 0

=0

[Kaelia]

Bless you, kind sirs. You have made tomorrow a better day.