20190118, 02:03  #155 
Jul 2003
wear a mask
3301_{8} Posts 
Let y = 1/x. Then x = 1/y. Limit as x goes to infinity is the same as limit as y goes to zero (from positive side). cos(x)/x = y*sin(1/y).
Could you use the squeeze theorem on this transformation of the original limit? 
20190118, 04:01  #156 
"Curtis"
Feb 2005
Riverside, CA
2^{2}·1,249 Posts 
squeeze it between 1/x and 1/x. Each of those limits is pretty easy to calculate!

20190118, 10:36  #157 
Bamboozled!
"πΊππ·π·π"
May 2003
Down not across
25312_{8} Posts 
It's fairly easy if you use exp(ix) = cos(x) + i sin(x) but perhaps you haven't been taught that yet.

20190118, 13:44  #158  
Feb 2017
Nowhere
5·997 Posts 
Quote:
Apply the squeeze theorem with 

20190119, 20:21  #159 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
This video helped a lot: https://www.youtube.com/watch?v=NPVDM3qFhFY
So we start with \(1 \leq \cos x \leq 1\), which is easy to visualize and is obviously true. We then multiply everything by \(\frac{1}{x}\) to get\[\dfrac{1}{x} \leq \dfrac{\cos x}{x} \leq \dfrac{1}{x}\]Now we evaluate each term as x approaches infinity. The limits of \(\dfrac{1}{x}\) and \(\dfrac{1}{x}\) evaluate to 0, leaving \[0 \leq \lim_{x\rightarrow \infty}\dfrac{\cos x}{x} \leq 0\]By the squeeze theorem/common sense, the remaining limit must equal 0. Last fiddled with by jvang on 20190119 at 20:21 Reason: typing is hard 
20190123, 02:28  #160 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Today we did a bit on finding the inverse of the derivatives of functions or something. A quick way to do it goes something like this: take the inverse of the original function, take the derivative of that, and take the reciprocal of it. \[\frac{d}{dx}f^{1}(x) = \dfrac{1}{f'(f^{1}(x))}\]Doesn't seem to difficult. The internet helped me figure out the basis for this formula, which seems to be simply using the chain rule on the definition of an inverse function: \[f(g(x)) = x\]Taking the derivative...\[f'(g(x)) \cdot g'(x) = 1\]Do some weird implicit differentiation stuff because it seems like it fits? \[f'(g(f(y))) \cdot g'(f(y)) = 1\]Cancel out some stuff in the first set of parentheses...\[f'(y) \cdot g'(f(y)) = 1\]\[f'(y) = \dfrac{1}{g'(f(y))}\]Seems legit...?

20190123, 17:20  #161  
Feb 2017
Nowhere
5·997 Posts 
Quote:
If your function is y = f(x), expressing the inverse function implicitly is easypeasy: x = f(y). Then, implicit differentiation tells us that 1 = f'(y) * y', or y' = 1/f'(y). Piece of cake! Except for one thing: Now you want to express f'(y) in terms of f(y), since x = f(y). So... the inverse tangent function may be expressed x = tan(y), so that (I checked here  you know the derivative of the tangent function) 1 = sec^{2}(y)*y', so that 1/sec^{2}(y) = y' Now  can you express sec^{2}(y) in terms of tan(y)? Of course you can! 

20190124, 01:18  #162  
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
444_{10} Posts 
Quote:
Second part: what do you mean by "in terms of tan(y)"? That sort of wording always messes me up... 

20190124, 06:23  #163 
"Jacob"
Sep 2006
Brussels, Belgium
6CF_{16} Posts 

20190124, 16:00  #164 
Feb 2017
Nowhere
5·997 Posts 

20190125, 02:32  #165 
veganjoy
"Joey"
Nov 2015
Middle of Nowhere,AR
2^{2}·3·37 Posts 
Much of the class today was on some sort of trip (FBLA club, they're always skipping school ), so we just went over some of this inverse derivative stuff. One particular problem resulted in a disagreement between the teacher and myself (me? I? Grammar is hard):
\(f(x) = x^3+2x+4\), and \(g(x)\) is its inverse. Find \(g(7)\) without finding the formula for \(g(x)\), then find \(g'(7)\). First, I worked with \(x^3+2x+4 = 7\), since the xinput for \(g(x)\) must be the youtput of \(f(x)\). This leads to 3 solutions to the equation, with the only real solution being 1. So we are dealing with two points; (1,7) for \(f(x)\) and (7,1) for \(g(x)\). Then we can use that weird theorem to plug in numbers. We take the reciprocal of the derivative of \(f(x)\), and plug in one of the coordinate values. Solving this should give the value of \(g'(7)\), but which value do you use? I think you're supposed to use the xvalue of \(f(x)\), 1, but our teacher thinks you're supposed to use 7. The results are either \(g'(7) = \dfrac{1}{5}\) for 1 or \(g'(7) = \dfrac{1}{149}\) for 7. It's been a couple of hours, and I can't remember my reasoning for picking 1, but I think that I'm correct. How do I navigate what's left of this problem? Reading this problem now, I'm kinda confused by its wording. I don't know how I got this far earlier today 
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