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\(\Huge \lim_{x \rightarrow 0}\frac{x-\tan x}{x^3}\)

can u use L'Hopital's ?

You sure can!
0-0 / 0 = 0
Wanna work on this together, inky?

Oh! Well, let's do it manually, then!
Ready?

First use of l'hopital's: \(\Huge \lim_{x \rightarrow 0}\frac{1-sec^2(x)}{3x^2}\)

Ready.

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'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.

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? :/

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).

-lim x->0 ( 2(tanx)*sec^2(x) ) / (6x)
-2/6 lim x -> 0 [tanxsec^2x]/[x]
Just one more l'hopital left!

Does this look wrong?

I got
tan(x)(2 sec^2(x) tan(x))+sec^4(x) for the numerator?

Sure.

Shall we post step by step or all steps?

same lol

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:
\(\Huge -\frac{1}{3}\lim_{x \rightarrow 0} \frac{\tan^2(x)}{x^2}\)

Yerp.

Basically, just pulled out the 3 from the denominator if it's not too clear.

Oh nice. Thinkin' alike. I like.

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)

Alrighty.

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. ;)

Lol - you know me don't you o.o

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

OH! Product rule, duh! Sorry.

lol I really have to stop using Tex, was 90% done then you figured it out xD

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

btw why is it justifiable that we can put the limit onto the exponent?

got it

ln(1+x)/x
(1/(x+1)*1)/1
1

me too.

Well bell rang, so I'll clean up and maybe see you later. Thanks for the help!