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inkyvoyd
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how does one evaluate the limit of:
Lim(x>0) (xtan(x))/(x^3)
 2 years ago
 2 years ago
inkyvoyd Group Title
how does one evaluate the limit of: Lim(x>0) (xtan(x))/(x^3)
 2 years ago
 2 years ago

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inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
\(\Huge \lim_{x \rightarrow 0}\frac{x\tan x}{x^3}\)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Wolfram says the answer is 1/3, but I'm more interested in a concise process  wolfram's evaluation steps are incredibly long and convoluted.
 2 years ago

hartnn Group TitleBest ResponseYou've already chosen the best response.0
can u use L'Hopital's ?
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
You sure can! 00 / 0 = 0 Wanna work on this together, inky?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
You can  but doing it the fast way does not involve l'hopital's  I tried it, and it's really not very efficient...
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Oh! Well, let's do it manually, then! Ready?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
First use of l'hopital's: \(\Huge \lim_{x \rightarrow 0}\frac{1sec^2(x)}{3x^2}\)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
use of trig identities: \(\Huge \lim_{x \rightarrow 0} \frac{\tan^2(x)}{3x^2}\)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Erm, that should be a negative.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
\(\Huge \lim_{x \rightarrow 0} \frac{\tan^2(x)}{3x^2}\)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
I am a tad bit embarrassed that I am admitting this, but I want to make sure to be on the safe side (it's been a while since I differentiated): You can legally add two halves of a differentiation to get the final answer, right? As in split it between differentiation of (x/x^3)  (tanx)/x^3 ? I ask this because it's been about a year since cal 1, and I'm in cal 2 right now and am pretty much only doing integrations, lol.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Yeah, you can split up the limit  the only problem is that we get two indeterminant forms instead of one, resulting in more chaos lol
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I'm wondering if I should go ahead and just use l'hopital's twice more...
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Yeah, we can definitely do that. I thought we were going to skip that lolz. Looking good so far.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
The only problem is that I'm pretty sure I'm that follows wolfram's steps.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
(tanx)^2 / (3x^2) Oh, it's ok, I think we'll be able to solve this eventually. Worst case scenario, we can fall back to that other method of manually doing the derivatives. Let's see where this gets us if we keep going.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Do you still remember the sandwich theorem?
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Oh dear, not at all. I was horrible at that one. Is this one of those instances where we need it? :/
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I'm pretty sure it would help, but I haven't gotten to it. My course (AP calc BC) just started on the "intuitive definition of limits", and want us to use a graphing calculator to evaluate limits by eyeballing it. My thoughts on that are  screw that, I didn't skim over a calc book last year to be spoonfed into doing limits the wrong way. So yeah, I pushed myself into an awkward situation with this limit.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
lim x>0 ( 2(tanx)*sec^2(x) ) / (6x) I like that your style, though. I admit I was lazy in cal 1, so that's why I'm so weak. But ironically, I'm a beast at Cal 2. I can integrate almost any reasonable expression and can find areas under a curve like it was a simple long division question. ;)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Woo! I think this next one will solve it for us!
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Well, assuming sec(0) gives us something besides undefined...
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Heck ya! We're good to go (it's 1).
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I think on this step wolf uses l'hopital's again but gets some factor of x on the bottom and screws everything up
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
lim x>0 ( 2(tanx)*sec^2(x) ) / (6x) 2/6 lim x > 0 [tanxsec^2x]/[x] Just one more l'hopital left!
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
We should be ok, though, because that x will turn into a 1. It's a matter of correctly chainruling the top, so correct me if I goof up (I will :P)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Lol this is probably going to be the worst derivative I've had in days  at least we have always have rules for those ;)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
denominator: (tanx)(secx)^2 (secx^2 * 2 (secx) * tanxsecx)/1 2 * (secx)^2 * tanx * secx 2 * (secx)^3 * tanx So... 2/6 lim x > 0 2 * (secx)^3 * tanx 4/6 lim x > 0 tanx(secx)^3 2/3 lim x > 0 tanx(sec^x)^3
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Does this look wrong?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I got tan(x)(2 sec^2(x) tan(x))+sec^4(x) for the numerator?
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Hmm.... let's work this out individually on paper and see if we get the same answer. I'm curious on this one and wanna get an answer. :P So wanna start from scratch and do it on paper and compare after every l'hopital?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Shall we post step by step or all steps?
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
I guess at the end of each l'hopital and then discuss if we differ. I guess I got too excited and went almost all the way. :P
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
After my first lhopital: lim (1  sec^2 x) / (3x^2)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
dang it... I mixed two steps together. Ignore the negative limit.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
First step \(\Huge \lim_{x \rightarrow 0}\frac{1\sec^2 x}{3x^2}\)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Mm, then both are the same.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Then I simplified into: 1/3 lim (tan^2 x)/(x^2) This should be legitimate. (No new lhopital yet)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Second step: \(\Huge \frac{1}{3}\lim_{x \rightarrow 0} \frac{\tan^2(x)}{x^2}\)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Basically, just pulled out the 3 from the denominator if it's not too clear.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Oh nice. Thinkin' alike. I like.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
So then after next lhopital: 1/3 lim [(tanx)(secx)^2]/x (I canceled a 2 in both the numerator and denominator)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
That's what I got  This whole TeX thing is messing with my typing speed lol
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Well I gotta finish this next step...
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
No prob; I can't complain, considering how ugly my writing is. :P
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Next up (let's do this step by step to be safe):
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
the tanx turns into (secx)^2
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Fair enough so far?
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I'm unreasonably bad at the chain rule, so I split it up into two product rules
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
(Keep a careful eye on my work; I might accidentally integrate instead :P)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Ah, ok. We should eventually arrive at the same answer, then. I'll keep typing out my steps in hopes that you might catch an obvious error by me.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
So for my (secx)^2, I dropped the 2 down in front of the secx: 2(sexc)^1 And then chainruled the inside: 2(secx)(secxtanx) I think we should have both gotten (secx)^2 * 2 * secx * secx * tanx for our numerators. Got anything similar? (Haven't simplified yet)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
yup that's what product rule yields.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
The denominator turned into a 1 for me (it was only an x before lhopitaling)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
btw, I'll start typing up my result with product rule
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Okie dokie; I'll start cleaning up the numerator.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
But only the numerator, no more limit crap and fractions lol, that stuff wastes so much time
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
(secx)^2 * 2 * secx * secx * tanx = secx * secx * 2 * secx * secx * tanx = 2 * (secx)^4 * tanx
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Aww man... but there's a two just begging to into the limit section. ;)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Oh nice. In one hour I'll have been up officially for 24 straight hours; roughly 15 of them spent on studying/math fun. :P
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
start out \(\Huge \tan(x)\sec^2(x)\) \(\Huge \tan(x)\frac{d\sec^2(x)}{dx}+\sec^4x\) \(\Huge \tan(x)(2\tan(x)\sec^2(x))+\sec^4x\) \(\Huge \tan(x)(2\tan(x)\sec^2(x))+\sec^4x\) Then we put in 0 for x, cause the denom is already reduced to one \(\Huge 0(\text{blahblahblahblahblah})+1^4\) \(\Huge 1\) Multiply by what's out of the limit \(\Huge \frac{1}{3}\) Rawrr.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Whoa. The site was dying/freezing on me there. Brings back memories of how this site used to be a year ago. lolz.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Lol  you know me don't you o.o
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
In real life? I doubt it. But we may have been friends on this site on the past (no offense or anything, but I don't recall your nickname).
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
omg the next problem I'm supposed to evaluate with the calculator is freakin insane.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
\(\Huge \lim_{x \rightarrow 0}(1+x)^{\frac{1}{x}}\)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
This one's either e or 1/e
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
One question, though: whenever I did mine, it came out to like 0, I think. Where'd your +sec^4 come from? I wanna learn that.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
OH! Product rule, duh! Sorry.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
lol I really have to stop using Tex, was 90% done then you figured it out xD
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Lolz, sorry. I was just curious since if I put in my x values on my numerator, I'd get 0. :P But all good now. Alright, let's do this! LEEEERRRROOOYYYYYYYy JENNNKINSSS!
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Alright, let's see if I've still got this in me:
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Lol, I'm outta time. But at least I have chicken. (not really)
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Start by doing e^ln(that crap)
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Well I got 3 more mins or so
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
^ Does the reasoning behind that make sense?
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
Aw man. Didn't realize you were timed. :(
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
I think 3 mins is enough though :o
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
and yes, it does make sense
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
alright. Got it to here: e to ln of (1+x)/x Try to solve that on paper really fast if you can. Get it to 0/0 or inf/inf to use lhopital.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
btw why is it justifiable that we can put the limit onto the exponent?
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
The limit is technically a variable so you can screw it it. You can multiply it, divide it, treat it as an exponent, etc.
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
So let's see: [ln(1+x)]/x e^ This is what I should have told you to write. And yes, the x is dividing the ln(1+x). This should be lhopitaable.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
ln(1+x)/x (1/(x+1)*1)/1 1
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
me too.
 2 years ago

inkyvoyd Group TitleBest ResponseYou've already chosen the best response.0
Well bell rang, so I'll clean up and maybe see you later. Thanks for the help!
 2 years ago

uber1337h4xx0r Group TitleBest ResponseYou've already chosen the best response.2
So e^1, I think. (Don't forget the e :P) Sounds good; have fun. I'm going to take a 4 hour lap before going to class I guess. :P I'll fan ya, maybe you can toss me some more probs; these were fun. See ya.
 2 years ago
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