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 2019-01-18, 02:03 #155 masser     Jul 2003 Behind BB 23·13·17 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?
 2019-01-18, 04:01 #156 VBCurtis     "Curtis" Feb 2005 Riverside, CA 507510 Posts squeeze it between -1/x and 1/x. Each of those limits is pretty easy to calculate!
 2019-01-18, 10:36 #157 xilman Bamboozled!     "πΊππ·π·π­" May 2003 Down not across 11·17·59 Posts It's fairly easy if you use exp(ix) = cos(x) + i sin(x) but perhaps you haven't been taught that yet.
2019-01-18, 13:44   #158
Dr Sardonicus

Feb 2017
Nowhere

5,147 Posts

Quote:
 Originally Posted by jvang There was a weird problem we worked with in class that no one was sure what to do with (teacher was as unsure as we were ). $\lim_{x\rightarrow \infty} \frac{\cos x}{x}$ Logically speaking, the answer looks like it'll be 0. With x going to positive infinity, the denominator will be infinitely larger than what the numerator will be, which is somewhere between -1 and 1.
Bingo. You have, assuming x is positive,

$\frac{-1}{x}\;\le\;\frac{\cos(x)}{x}\;\le\;\frac{1}{x}$

Apply the squeeze theorem with $x\;\rightarrow\;+\infty$

 2019-01-19, 20:21 #159 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 6748 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 2019-01-19 at 20:21 Reason: typing is hard
 2019-01-23, 02:28 #160 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 1101111002 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...?
2019-01-23, 17:20   #161
Dr Sardonicus

Feb 2017
Nowhere

514710 Posts

Quote:
 Originally Posted by jvang Do some weird implicit differentiation stuff because it seems like it fits?
Implicit differentiation is great for inverse functions -- provided you can write the derivative in terms of the original function.

y = f(x),

expressing the inverse function implicitly is easy-peasy:

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 = sec2(y)*y', so that

1/sec2(y) = y'

Now -- can you express sec2(y) in terms of tan(y)? Of course you can!

2019-01-24, 01:18   #162
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22×3×37 Posts

Quote:
 Originally Posted by Dr Sardonicus Implicit differentiation is great for inverse functions -- provided you can write the derivative in terms of the original function. If your function is y = f(x), expressing the inverse function implicitly is easy-peasy: x = f(y). (...) Now -- can you express sec2(y) in terms of tan(y)? Of course you can!
First part: that looks a lot like what we did in algebra class. It makes a bit more sense in this application/context!

Second part: what do you mean by "in terms of tan(y)"? That sort of wording always messes me up...

2019-01-24, 06:23   #163
S485122

"Jacob"
Sep 2006
Brussels, Belgium

22·439 Posts

Quote:
 Originally Posted by Dr Sardonicus ... If your function is y = f(x), expressing the inverse function implicitly is easy-peasy: x = f(y). ...
Shouldn't this be x=f -1(y). Only when f(x)=x does y=f(x) imply x=(f(y) for any x where f(x) is defined and invertible...

Jacob

2019-01-24, 16:00   #164
Dr Sardonicus

Feb 2017
Nowhere

10100000110112 Posts

Quote:
 Originally Posted by jvang Second part: what do you mean by "in terms of tan(y)"? That sort of wording always messes me up...
There's a relation between sec2 and tan2 that follows from the well-known relation between sin2 and cos2...

 2019-01-25, 02:32 #165 jvang veganjoy     "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 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 x-input for $$g(x)$$ must be the y-output 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 x-value 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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