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Old 2018-09-25, 08:24   #45
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Our teacher said that the limit does not exist here, but I asked if it would be incorrect to answer the problem with \(\lim_{x\rightarrow -1}f(x) = \infty\). Is this correct, or does the limit not exist?
You are right, but saying that the limit is infinity is a different type of limit with a different definition.
(If your class has been taught the first definition already but not yet the second one then it would be logical to say that the limit does not exist.)
In Spivak's Calculus book, the second definition is in one of the exercises in the chapter on limits.
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We also covered some rules regarding limits, which basically seem like treating \(\lim\) as a function...However, it seems like we could just plug in whatever x-value we're looking for would work in most cases?
If a function f is continuous at a point c then it follows that
\[ \lim_{x\rightarrow c}f(x)=f(c). \]
In that case, you can just replace the x's with c's, as you correctly point out.
But a function can have a limit as x tends to c without being continuous there, or without even being defined at the point c itself!
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Man, my limit notation is still messed up! How was it working earlier in the thread?
If you want LaTeX to put the arrow under the word "lim", display the mathematics on a separate line (using escaped square brackets).
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Old 2018-09-26, 23:50   #46
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We covered how to work with some functions with discontinuities, such as: \[\lim_{x\rightarrow 4}\dfrac{x^2-16}{x-4}\]For this one, substitution gives us 0/0, which is undefined. However, we can do some (kinda) fancy algebra tricks to see if a limit exists...\[\dfrac{x^2-16}{x-4}=\dfrac{(x+4)(x-4)}{x-4}=x+4\]Factoring out the \(\dfrac{x-4}{x-4}\) lets us evaluate the limit of the resulting function, while keeping in mind that the domain of the original function is undefined at \(x=4\). Thus:\[\lim_{x\rightarrow 4}(x+4) = 8\]This works for functions that have removable discontinuities (the book we are using mentions undefined expressions such as \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(\infty \cdot 0\), and \(\infty - \infty\). Other types, such as \(1^{\infty}\) and \(\infty^0\) are covered in a later chapter.).

Trying to factor out terms of a function that has a vertical asymptote does not result in a continuous function; that type of discontinuity is an infinite discontinuity, and we haven't learned about how to work with those algebraically yet. The book states that we can still describe the one-sided limits to vertical asymptotes, but we didn't mention this in class. We really didn't look at the book, and missed out on a lot of important concepts. Oh well
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Old 2018-09-27, 08:47   #47
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For this one, substitution gives us 0/0, which is undefined. However, we can do some (kinda) fancy algebra tricks to see if a limit exists...
There's another trick which works here and in many other cases too, so it's worth remembering:
if taking the limits of the top and bottom of a fraction would give you 0/0, you can differentiate top and bottom separately and then take their limits instead.
In your example, the derivative of \(x^2-16\) is \(2x\) and the derivative of \(x-4\) is 1, so
\[ \lim_{x\rightarrow 4}\frac{x^2-16}{x-4}=\lim_{x\rightarrow 4}\frac{2x}{1}=8. \]
This is known as "de l'Hôpital's rule".
(It's difficult to find in the index of books as you never know whether it will be under D, L or H!)
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Old 2018-09-27, 13:27   #48
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if taking the limits of the top and bottom of a fraction would give you 0/0, you can differentiate top and bottom separately and then take their limits instead.
In your example,
...
This is known as "de l'Hôpital's rule".
(It's difficult to find in the index of books as you never know whether it will be under D, L or H!)
Does our young learner know what differentiate means yet? One can't really grok derivatives until limits and continuity are understood, right?
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Old 2018-09-28, 23:56   #49
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We have yet to cover L'Hopital's rule or derivatives, but they are coming soon™! (Probably 2 or 3 weeks from now...)

We're still covering how to substitute values into functions to determine limits, and factoring out removable discontinuities. Here's a weird one we worked with today:\[\lim_{x\rightarrow 2}\dfrac{2^{2x}+2^x-20}{2^x-4}\]\[\lim_{x\rightarrow 2}\dfrac{(2^x)^2+2^x-20}{2^x-4}\]Using exponent rules, we can see how to factor this (treating \(2^x\) as \(x\) or some other variable):\[\lim_{x\rightarrow 2}\dfrac{(2^x+5)(2^x-4)}{2^x-4}\]Hm, I vaguely remember some Latex code for AoPS that did diagonal strikethroughs...

This simplifies to:\[\lim_{x\rightarrow 2}(2^x+5)=2^2+5=9\]Another neat one:\[\lim_{x\rightarrow \frac{\pi}{4}}\dfrac{\sin x-\cos x}{\tan x-1}\]Using identities is one of the few parts of trigonometry that is neat...\[\lim_{x\rightarrow \frac{\pi}{4}}\dfrac{\sin x-\cos x}{\frac{\sin x}{\cos x}-1}\]\[\lim_{x\rightarrow \frac{\pi}{4}}\dfrac{\sin x-\cos x}{\frac{\sin x - \cos x}{\cos x}}\]\[\lim_{x\rightarrow \frac{\pi}{4}}\dfrac{\sin x - \cos x}{1} \cdot \dfrac{\cos x}{\sin x - \cos x}\]Cancel out the first numerator and second denominator...\[\lim_{x\rightarrow \frac{\pi}{4}}\cos x = \cos \frac{\pi}{4} = \frac{\sqrt{2}}{2}\]
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Old 2018-10-03, 01:10   #50
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I was doing some problems and had a weird thought. Let's say we have this:\[\dfrac{x-1}{\sqrt{x}+1}\]And we would like to simplify it. One way is to remove a factor from the top and bottom:\[\dfrac{(\sqrt{x}+1)(\sqrt{x}-1)}{\sqrt{x}+1}\]\[\sqrt{x} - 1\]Another way that makes more sense to some people is to multiply the fraction by the conjugate of the denominator:\[\dfrac{x-1}{\sqrt{x}+1} \cdot \dfrac{\sqrt{x}-1}{\sqrt{x}-1}\]Which makes the common factors of the numerator and denominator more clear. However, what happens if we multiply by the denominator itself:\[\dfrac{x-1}{\sqrt{x}+1} \cdot \dfrac{\sqrt{x}+1}{\sqrt{x}+1}\]I was thinking that this is worth looking into (cancel out the denominator), but now that I've typed it out it looks much less intuitive...

It's pretty wrong, isn't it
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Old 2018-10-03, 04:40   #51
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What happens to the new denominator if you multiply those factors out? Does the radical disappear like it does in your first two methods?
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Old 2018-10-03, 23:35   #52
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What happens to the new denominator if you multiply those factors out? Does the radical disappear like it does in your first two methods?
It doesn't

I wasn't thinking this through very well. We had just did some trig stuff where we canceled out denominators in a different way that seemed similar but really wasn't
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Old 2018-10-12, 14:44   #53
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However, what happens if we multiply by the denominator itself:\[\dfrac{x-1}{\sqrt{x}+1} \cdot \dfrac{\sqrt{x}+1}{\sqrt{x}+1}\]I was thinking that this is worth looking into (cancel out the denominator), but now that I've typed it out it looks much less intuitive...

It's pretty wrong, isn't it
It's not wrong. It's just not very useful. But it's good to be in the habit of trying different things and seeing what works.
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Old 2018-10-16, 01:04   #54
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I'm under the impression that we aren't actually going to learn the formal epsilon-delta definition of a limit, given that we're jumping to the squeeze theorem without a proof of how it or limits work

The squeeze theorem is neat, but we're not really using it. We just use that fact that \[\lim_{x\rightarrow 0}\dfrac{\sin x}{x} = 1\]Or some other squeeze theorem identity that equals 1, then solve an indeterminate trigonometric equation by substituting the identities in. An example:\[\lim_{x\rightarrow 0}\dfrac{\sin 3x}{x} \cdot \dfrac{3}{3} = 3 \lim_{x\rightarrow 0}\dfrac{\sin 3x}{3x}\]We multiply by 3/3, and distribute the numerator and denominator separately. We know that, in this special case where the denominator and the argument (is that the term?) of the sine function are equal, that the limit is 1, so we have \[3\lim_{x\rightarrow 0}\dfrac{\sin 3x}{3x} = 3 \cdot 1 = 3\]Another thing we are doing is finding limits as x tends towards infinity. One case we worked with was \[\lim_{x\rightarrow \infty}\dfrac{2x^3-4x}{3x^4+2x^2}\]Once again we are using identities, this one being that \[\lim_{x\rightarrow \infty}\dfrac{1}{x} = 0\] for any constant numerator and coefficient of x. The way to solve these problems is to multiply the limit equation by the reciprocal of the highest degree term, in a way that it still equals 1: \[\dfrac{\frac{1}{x^4}}{\frac{1}{x^4}}\]Then distribute until you have constant terms and terms that can be replaced with the identity. In this particular case, I figured that we could factor out 1/x from the original equation and use limit laws and our identity: \[\lim_{x\rightarrow \infty}\dfrac{2x^3-4x}{3x^4+2x^2} = \lim_{x\rightarrow \infty}\dfrac{2x^3-4x}{3x^3+2x} \cdot \lim_{x\rightarrow \infty}\dfrac{1}{x}\]Which would simplify to \[\lim_{x\rightarrow \infty}\dfrac{2x^3-4x}{3x^3+2x} \cdot 0 = 0\]Which is correct. I'd guess that this works as long as there is a common x to each term in the denominator?
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Old 2018-10-16, 04:54   #55
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On your last line, couldn't you factor an x out of the top instead and make the same argument about an expression times infinity? What would that limit be?

Your last line has 0 times an unknown quantity. What if that unknown quantity goes to inifinity while your known term goes to zero? Note that, even if you know what infinity times zero is (you don't), you have a quantity that might be "going to infinity" multiplied by a quantity that is definitely "going to zero". The problem is that you don't know whether one is growing faster than the other is shrinking, so it's not so wise to break up limits in the way you did (unless you can demonstrate that the "unknown" part goes to a constant, which in this example it does via the divide-by-biggest-power-on-bottom trick you already discussed).
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