 mersenneforum.org Introductory Calculus Discussion Thread
 Register FAQ Search Today's Posts Mark Forums Read  2018-08-31, 23:32   #23
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

1BC16 Posts Quote:
 Originally Posted by VBCurtis We can use separate values, when a one-sided limit is requested. We can say "the limit from the left is N", or "the limit from the right is M". But, when a limit doesn't specify one-sided-ness, our answer must be valid from either side. So, if the limits from each side exist but do not match, we have valid answers for each one-sided limit, but DNE for the regular (two-sided) limit.
Ok, that makes sense. So is there a difference in the notation used for two-sided and one-sided limits? Or would the question usually clear that up?

Quote:
 Originally Posted by LaurV There are many ways, one is to use atop or use a matrix to align them
Hmmm, looks more straightforward than I figured it might be. Thanks!

Of course, now I'll have to figure out how to make the brace properly sized Edit: Holy crap; after reading all the Google results on how normal Latex doesn't really support it (people want to import packages, although there was some other really weird vanilla way), I did it!

$$f(x)=\bigg{\{}\begin{array}{cc} x-1 & : & x<1 \\ x^3 & : & x>1 \end{array}$$ Last fiddled with by jvang on 2018-08-31 at 23:49 Reason: typing is hard   2018-09-01, 09:19   #24
Nick

Dec 2012
The Netherlands

174010 Posts Quote:
 Originally Posted by jvang So is there a difference in the notation used for two-sided and one-sided limits?
One notation is to write, for example,
$\lim_{x\rightarrow 0-}\frac{x}{|x|}=-1; \lim_{x\rightarrow 0+}\frac{x}{|x|}=1$
where "0-" means "0 from below" and "0+" means "0 from above".

Some people find that notation confusing, however, and prefer to write
$\lim_{x\uparrow 0}\frac{x}{|x|}=-1; \lim_{x\downarrow 0}\frac{x}{|x|}=1.$   2018-09-02, 14:56   #25
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22×3×37 Posts Quote:
 Originally Posted by Nick One notation is to write, for example, $\lim_{x\rightarrow 0-}\frac{x}{|x|}=-1; \lim_{x\rightarrow 0+}\frac{x}{|x|}=1$ where "0-" means "0 from below" and "0+" means "0 from above". Some people find that notation confusing, however, and prefer to write $\lim_{x\uparrow 0}\frac{x}{|x|}=-1; \lim_{x\downarrow 0}\frac{x}{|x|}=1.$
I think that I've seen$\lim_{x\rightarrow 0}\frac{x}{|x|}=-1; \lim_{x\leftarrow 0}\frac{x}{|x|}=1$with a left arrow, but that looks weird given the other ways you've shown, and I might be thinking of something else (I can't find any examples of this format).   2018-09-02, 16:56 #26 LaurV Romulan Interpreter   Jun 2011 Thailand 24×13×47 Posts We used to write diagonal arrows (pointing up-right or down-right) but the +/- notation was also common, and used to pronounce "x grows to 0" or "raises to 0", etc.   2018-09-15, 02:59 #27 jvang veganjoy   "Joey" Nov 2015 Middle of Nowhere,AR 22×3×37 Posts We've been going over basic trigonometry over the past couple of days (not extremely basic; mostly identities and unit circle usage). I am actually really awful at trigonometry; it seems like nothing but rote memorization and is one of the most boring, though most widespread, topics in math that I have learned. Here's a couple of problems that I was struggling with (not exactly homework but maybe belongs in the Homework Help thread? ):$\cot x = \dfrac 1 3$I was not very sure how to work this out; I'm not actually really sure how you'd solve something like $$\sin x = \dfrac{\sqrt{3}}{2}$$. We aren't allowed to use a calculator, so all I have is what I've memorized from a unit circle, which tells me that the sine value is $$\dfrac{\sqrt{3}}{2}$$ at $$\dfrac{\pi}{3}$$ radians. To be completely honest, I can barely follow the basic trigonometric functions in regards to a right triangle; I have to use a crappy mnemonic to remember what goes where because I haven't found an intuitive way to think of them... Additionally, our teacher hasn't covered the notation that accounts for infinite $$x$$-values in trigonometric equations, such as the solutions (or at least what I think they are) to $$\sin x = \dfrac{\sqrt{3}}{2}$$ actually being $$\dfrac{\pi}{3} + n(2\pi)$$ and $$\dfrac{5\pi}{3} + n(2\pi)$$ for an integer $$n$$. This notation is how I learned it at ASMSA (which is probably the standard), and it's confusing when my teacher points out that she'd rather I restrict my domain to $$[0,2\pi]$$ (I've been getting my domains and unit circle range mixed up, but I think I'm getting better about it). In fact, while writing this post, I messed up the sine equation that I was using as an example. Ugh! One that I think was either written incorrectly or is beyond what I/we have covered:$2 \cos x = 13 \sin x + 5$It's likely that there is an identity that you could substitute in, but I am unaware of such a solution. The best idea I could think of was to divide by $$\cos x$$ to attempt to isolate the trigonometric function, but the constant term is in the way...   2018-09-15, 06:32   #28
xilman
Bamboozled!

"𒉺𒌌𒇷𒆷𒀭"
May 2003
Down not across

101010110001012 Posts Quote:
 Originally Posted by jvang We've been going over basic trigonometry over the past couple of days (not extremely basic; mostly identities and unit circle usage). I am actually really awful at trigonometry; it seems like nothing but rote memorization and is one of the most boring, though most widespread, topics in math that I have learned. Here's a couple of problems that I was struggling with (not exactly homework but maybe belongs in the Homework Help thread? ):$\cot x = \dfrac 1 3$I was not very sure how to work this out; I'm not actually really sure how you'd solve something like $$\sin x = \dfrac{\sqrt{3}}{2}$$. We aren't allowed to use a calculator, so all I have is what I've memorized from a unit circle, which tells me that the sine value is $$\dfrac{\sqrt{3}}{2}$$ at $$\dfrac{\pi}{3}$$ radians. To be completely honest, I can barely follow the basic trigonometric functions in regards to a right triangle; I have to use a crappy mnemonic to remember what goes where because I haven't found an intuitive way to think of them... Additionally, our teacher hasn't covered the notation that accounts for infinite $$x$$-values in trigonometric equations, such as the solutions (or at least what I think they are) to $$\sin x = \dfrac{\sqrt{3}}{2}$$ actually being $$\dfrac{\pi}{3} + n(2\pi)$$ and $$\dfrac{5\pi}{3} + n(2\pi)$$ for an integer $$n$$. This notation is how I learned it at ASMSA (which is probably the standard), and it's confusing when my teacher points out that she'd rather I restrict my domain to $$[0,2\pi]$$ (I've been getting my domains and unit circle range mixed up, but I think I'm getting better about it). In fact, while writing this post, I messed up the sine equation that I was using as an example. Ugh! One that I think was either written incorrectly or is beyond what I/we have covered:$2 \cos x = 13 \sin x + 5$It's likely that there is an identity that you could substitute in, but I am unaware of such a solution. The best idea I could think of was to divide by $$\cos x$$ to attempt to isolate the trigonometric function, but the constant term is in the way...
You've already posted enough show me that you know enough to solve that latter equation.

First, let's establish an identity between sin and cos which you've likely met already. Draw a right-angled triangle with a hypotenuse of length 1 and one angle x. The other two sides have length cos x and sin x. (Your mnemonic may well have been Soon Old Horses Can All Have Tons Of Apples (sin = opposite/hypotenuse, cos= adjacent/hypotenuse) and use Pythagorus' Theorem.

Then use that identity to simplify your problematic equation. Hint: you should end up with a quadratic.   2018-09-15, 23:26   #29
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22·3·37 Posts Quote:
 Originally Posted by xilman You've already posted enough show me that you know enough to solve that latter equation. First, let's establish an identity between sin and cos which you've likely met already. Draw a right-angled triangle with a hypotenuse of length 1 and one angle x. The other two sides have length cos x and sin x. (Your mnemonic may well have been Soon Old Horses Can All Have Tons Of Apples (sin = opposite/hypotenuse, cos= adjacent/hypotenuse) and use Pythagorus' Theorem. Then use that identity to simplify your problematic equation. Hint: you should end up with a quadratic.
Yes, the mnemonic is some weird form of SOHCAHTOA. The kids at my school have a ton of made-up mnemonics for everything, but the most common is distributing ("FOIL", something very weird for a linear polynomial raised to the third power, and other extremely specific applications.)

Hmm...here's a picture that I think shows what you described: To fill in the other angle (opposite the side with length = $$\cos x$$), I think that it is equal to $$\dfrac{\pi}{2} - x$$. Wait...this didn't matter $$a^2+b^2 = c^2$$

$$(\sin x)^2+(\cos x)^2=1^2$$

$$\sin^2 x + \cos^2 x = 1$$

This looks very familiar, and I'm thankful that you have given me a logical representation of how the identity works! So we have:

$$2 \cos x = 13 \sin x + 5$$

Hmmm, how to use our identity? Do we square everything?

$$4\cos^2x = 169\sin^2x + 130\sin x + 25$$

$$4(1-\sin^2x) = 169\sin^2x + 130 \sin x + 25$$

$$4-4\sin^2x = 169\sin^2x + 130 \sin x + 25$$

$$173\sin^2x + 130 \sin x + 21 = 0$$

I suppose we apply the quadratic formula...?

$$x = \dfrac{-b\pm \sqrt{b^2-4ac}}{2a}$$

$$x = \dfrac{-(130)\pm \sqrt{130^2-4(173)(21)}}{2(173)}$$

$$x = \dfrac{-130\pm \sqrt{16,900-14,532}}{346}$$

$$x = \dfrac{-130\pm \sqrt{2,368}}{346}$$

$$x = \dfrac{-130\pm 8 \sqrt{37}}{346}$$

$$x = \dfrac{-65\pm 4 \sqrt{37}}{173}$$

Then I'd need to find the sine of both $$x$$s? I'm under the impression that there is a typo somewhere in the question (and my answer!)...

Last fiddled with by jvang on 2018-09-15 at 23:31 Reason: typing is hard   2018-09-16, 02:12   #30
VBCurtis

"Curtis"
Feb 2005
Riverside, CA

27·3·13 Posts Quote:
 Originally Posted by jvang Then I'd need to find the sine of both $$x$$s? I'm under the impression that there is a typo somewhere in the question (and my answer!)...
No, solving an equation means you want to find x, not sinx. There may well be a typo in the question, but your solution looks fine.   2018-09-16, 03:11   #31
jvang
veganjoy

"Joey"
Nov 2015
Middle of Nowhere,AR

22×3×37 Posts Quote:
 Originally Posted by VBCurtis No, solving an equation means you want to find x, not sinx. There may well be a typo in the question, but your solution looks fine.
Right, but wouldn’t I need to find the inverse sine of x since sin is applied in the equation?

In any case it’s not often that the solutions to what our teacher gives are as convoluted as this one, and she makes little typos more often than not. I’d think that the 2cos x was supposed to be a 2cos^2 x, or something similar.   2018-09-16, 04:03 #32 VBCurtis   "Curtis" Feb 2005 Riverside, CA 27×3×13 Posts Sorry, yes, I missed that you substituted x for sinx when you did the quadratic formula. You're quite correct that you'd want sine-inverse of each result from quadratic. While I am shamed by not noticing the substitution, it's generally not good practice to substitute using the same variable; using u = sinx rather than x = sinx avoids such confusion (in the student, or his proofreader!).   2018-09-18, 23:02 #33 jvang veganjoy   "Joey" Nov 2015 Middle of Nowhere,AR 22·3·37 Posts We have finally started on actual calculus topics! We learned about three types of limits: two-sided, one-sided, and infinite. A two-sided limit approaches the same value from both sides (from greater and lesser values of $$x$$), a one-sided limit approaches a value from only one side, and an infinite limit is one where the value of $$y$$ approaches infinity near a certain point (can be one- or two-sided). Here's some relevant syntax: Two-sided: $$\lim_{x\rightarrow 0}f(x) = 1$$ One-sided from greater values: $$\lim_{x\rightarrow 0+}f(x) = 1$$ One-sided from lesser values: $$\lim_{x\rightarrow 0-}f(x) = 1$$ Two-sided to positive infinity: $$\lim_{x\rightarrow 0}f(x) = \infty$$ Uh oh, the limit notation isn't rendering properly! What did I do wrong... The teacher said that the infinite limits correspond to vertical asymptotes, but would the other forms of limits technically be vertical asymptotes, too? Maybe my definition of asymptotes is misguided We also covered graphs that oscillate between positive and negative while approaching the limit value, which means that we couldn't find the limit. An example was $$f(x) = \sin \dfrac{\pi}{x}$$. Is there a way to find the limit of this function? We won't be covering the algebraic methods of finding limits for another two weeks; this lesson was to introduce the idea of a limit in a visual way. Guess this is going to take a while...   Thread Tools Show Printable Version Email this Page Similar Threads Thread Thread Starter Forum Replies Last Post fivemack Lounge 12 2018-08-11 07:51 jvang Lounge 27 2018-08-10 13:32 Rincewind Five or Bust - The Dual Sierpinski Problem 57 2011-02-06 21:53 philmoore Five or Bust - The Dual Sierpinski Problem 83 2010-09-25 10:20 philmoore Five or Bust - The Dual Sierpinski Problem 66 2010-02-10 14:34

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