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inkyvoyd
how does one evaluate the limit of: Lim(x->0) (x-tan(x))/(x^3)
\(\Huge \lim_{x \rightarrow 0}\frac{x-\tan x}{x^3}\)
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.
You sure can! 0-0 / 0 = 0 Wanna work on this together, inky?
You can - but doing it the fast way does not involve l'hopital's - I tried it, and it's really not very efficient...
Oh! Well, let's do it manually, then! Ready?
First use of l'hopital's: \(\Huge \lim_{x \rightarrow 0}\frac{1-sec^2(x)}{3x^2}\)
use of trig identities: \(\Huge \lim_{x \rightarrow 0} \frac{\tan^2(x)}{3x^2}\)
Erm, that should be a negative.
\(\Huge -\lim_{x \rightarrow 0} \frac{\tan^2(x)}{3x^2}\)
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.
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
I'm wondering if I should go ahead and just use l'hopital's twice more...
Yeah, we can definitely do that. I thought we were going to skip that lolz. Looking good so far.
The only problem is that I'm pretty sure I'm that follows wolfram's steps.
(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.
Do you still remember the sandwich theorem?
Oh dear, not at all. I was horrible at that one. Is this one of those instances where we need it? :/
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.
-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. ;)
Woo! I think this next one will solve it for us!
Well, assuming sec(0) gives us something besides undefined...
Heck ya! We're good to go (it's 1).
I think on this step wolf uses l'hopital's again but gets some factor of x on the bottom and screws everything up
-lim x->0 ( 2(tanx)*sec^2(x) ) / (6x) -2/6 lim x -> 0 [tanxsec^2x]/[x] Just one more l'hopital left!
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)
Lol this is probably going to be the worst derivative I've had in days - at least we have always have rules for those ;)
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
Does this look wrong?
I got tan(x)(2 sec^2(x) tan(x))+sec^4(x) for the numerator?
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?
Shall we post step by step or all steps?
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
After my first lhopital: -lim (1 - sec^2 x) / (3x^2)
dang it... I mixed two steps together. Ignore the negative limit.
First step \(\Huge \lim_{x \rightarrow 0}\frac{1-\sec^2 x}{3x^2}\)
Mm, then both are the same.
Then I simplified into: -1/3 lim (tan^2 x)/(x^2) This should be legitimate. (No new lhopital yet)
Second step: \(\Huge -\frac{1}{3}\lim_{x \rightarrow 0} \frac{\tan^2(x)}{x^2}\)
Basically, just pulled out the 3 from the denominator if it's not too clear.
Oh nice. Thinkin' alike. I like.
So then after next lhopital: -1/3 lim [(tanx)(secx)^2]/x (I canceled a 2 in both the numerator and denominator)
That's what I got - This whole TeX thing is messing with my typing speed lol
Well I gotta finish this next step...
No prob; I can't complain, considering how ugly my writing is. :P
Next up (let's do this step by step to be safe):
the tanx turns into (secx)^2
Fair enough so far?
I'm unreasonably bad at the chain rule, so I split it up into two product rules
(Keep a careful eye on my work; I might accidentally integrate instead :P)
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.
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)
yup that's what product rule yields.
The denominator turned into a 1 for me (it was only an x before lhopitaling)
btw, I'll start typing up my result with product rule
Okie dokie; I'll start cleaning up the numerator.
But only the numerator, no more limit crap and fractions lol, that stuff wastes so much time
(secx)^2 * 2 * secx * secx * tanx = secx * secx * 2 * secx * secx * tanx = 2 * (secx)^4 * tanx
Aww man... but there's a two just begging to into the limit section. ;)
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
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.
Whoa. The site was dying/freezing on me there. Brings back memories of how this site used to be a year ago. lolz.
Lol - you know me don't you o.o
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).
omg the next problem I'm supposed to evaluate with the calculator is freakin insane.
\(\Huge \lim_{x \rightarrow 0}(1+x)^{\frac{1}{x}}\)
This one's either e or 1/e
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.
OH! Product rule, duh! Sorry.
lol I really have to stop using Tex, was 90% done then you figured it out xD
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!
Alright, let's see if I've still got this in me:
Lol, I'm outta time. But at least I have chicken. (not really)
Start by doing e^ln(that crap)
Well I got 3 more mins or so
^ Does the reasoning behind that make sense?
Aw man. Didn't realize you were timed. :(
I think 3 mins is enough though :o
and yes, it does make sense
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.
btw why is it justifiable that we can put the limit onto the exponent?
The limit is technically a variable so you can screw it it. You can multiply it, divide it, treat it as an exponent, etc.
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.
ln(1+x)/x (1/(x+1)*1)/1 1
Well bell rang, so I'll clean up and maybe see you later. Thanks for the help!
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.