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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You sure can! 00 / 0 = 0 Wanna work on this together, inky?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh! Well, let's do it manually, then! Ready?

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

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah, we can definitely do that. I thought we were going to skip that lolz. Looking good so far.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh dear, not at all. I was horrible at that one. Is this one of those instances where we need it? :/

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0I'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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0lim 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. ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Woo! I think this next one will solve it for us!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Well, assuming sec(0) gives us something besides undefined...

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Heck ya! We're good to go (it's 1).

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We 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)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0denominator: (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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Does this look wrong?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Hmm.... 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?

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0After my first lhopital: lim (1  sec^2 x) / (3x^2)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dang it... I mixed two steps together. Ignore the negative limit.

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

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0Mm, then both are the same.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Then I simplified into: 1/3 lim (tan^2 x)/(x^2) This should be legitimate. (No new lhopital yet)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Basically, just pulled out the 3 from the denominator if it's not too clear.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh nice. Thinkin' alike. I like.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So then after next lhopital: 1/3 lim [(tanx)(secx)^2]/x (I canceled a 2 in both the numerator and denominator)

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No prob; I can't complain, considering how ugly my writing is. :P

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Next up (let's do this step by step to be safe):

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the tanx turns into (secx)^2

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(Keep a careful eye on my work; I might accidentally integrate instead :P)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Ah, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The denominator turned into a 1 for me (it was only an x before lhopitaling)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okie dokie; I'll start cleaning up the numerator.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0(secx)^2 * 2 * secx * secx * tanx = secx * secx * 2 * secx * secx * tanx = 2 * (secx)^4 * tanx

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Aww man... but there's a two just begging to into the limit section. ;)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oh nice. In one hour I'll have been up officially for 24 straight hours; roughly 15 of them spent on studying/math fun. :P

inkyvoyd
 4 years ago
Best ResponseYou've already chosen the best response.0start 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Whoa. The site was dying/freezing on me there. Brings back memories of how this site used to be a year ago. lolz.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0In 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).

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

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0One 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OH! Product rule, duh! Sorry.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Lolz, 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!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, let's see if I've still got this in me:

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Start by doing e^ln(that crap)

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0^ Does the reasoning behind that make sense?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Aw man. Didn't realize you were timed. :(

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0alright. 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.

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0The limit is technically a variable so you can screw it it. You can multiply it, divide it, treat it as an exponent, etc.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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.

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

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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.
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