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Old 2019-01-18, 02:03   #155
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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?
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Old 2019-01-18, 04:01   #156
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squeeze it between -1/x and 1/x. Each of those limits is pretty easy to calculate!
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Old 2019-01-18, 10:36   #157
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It's fairly easy if you use exp(ix) = cos(x) + i sin(x) but perhaps you haven't been taught that yet.
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Old 2019-01-18, 13:44   #158
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Quote:
Originally Posted by jvang View Post
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
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Old 2019-01-19, 20:21   #159
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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
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Old 2019-01-23, 02:28   #160
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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...?
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Old 2019-01-23, 17:20   #161
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Quote:
Originally Posted by jvang View Post
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.

If your function is

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!
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Old 2019-01-24, 01:18   #162
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Quote:
Originally Posted by Dr Sardonicus View Post
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...
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Old 2019-01-24, 06:23   #163
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Quote:
Originally Posted by Dr Sardonicus View Post
...
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
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Old 2019-01-24, 16:00   #164
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Quote:
Originally Posted by jvang View Post
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...
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Old 2019-01-25, 02:32   #165
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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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