Hello, is there a way to make wolfram calculate limits without

Consider the function

[eqn] f(x) = \begin{cases}
\frac{1 - \cos(x)}{x} & \text{ if } x \neq 0 \\
0 & \text{ if } x= 0
\end{cases} [/eqn]

Then

[eqn] \lim_{x \to 0} \frac{x^2}{1 - \cos(x)} = \frac{1}{ \lim_{x \to 0} \frac{f(x) - f(0)}{x}} = \frac{1}{f'(0)} = 2[/eqn]

>>8514457
How does this work? Sorry for not seeing it :(

>>8514457
I assume he isn't allowed to use l'Hopital yet because the derivative hasn't been formally defined in class.

>>8514486
Yep

>>8514445
Damn I just realized I completely forgot how to calculate limits.

don't use wolfram to do your homework for you. work on your problems sets yourself.
hint: multiply numerator and denominator by 1+cos(x)

>>8514507
i dunno it feels like such a waste of time doing it in an inefficient way just because "you're not allowed".
just write the proofs and you're good.
doesn't matter if you're allowed when it's formaly correct.

>>8514622
You're a genius
I did this just so I could see if I could do it

so multiply top and bottom by 1+cosx you get:

limx->0 x^2(1+cosx)/(1-cos^2x)
= limx->0 x^2(1+cosx)/sin^2x

then since the limit of products is the product of limits we have

= limx->0 x^2/sin^2x * limx->0 (1+cosx)
= limx->0 (x/sinx)^2 * (1+cos(0))
= (limx->0 x/sinx)^2 * 2

by the squeeze theorem we have the first limit equals 1 so:
= 1*2 = 2
QED

>>8514638
It's not "inefficient", it's the most correct way to do the problem in a sense. If you don't understand how L'Hopital came up with his rule then you don't really understand it. Math isn't about tricks to help you do problems, it's about compounding truths on each other to find higher truths.

>>8514736
>If you don't understand how L'Hopital came up with his rule then you don't really understand it.

What if I had to prove l'hopital's rule for my calculus class?

I am a brainlet and used infinitesimals in my proof! but it is still a proof! I know exactly why it works.

>>8514764
jej
was lhopital first proven using our modern (Cauchy's?) definition of a limit tho?

>>8514777
No idea. In the proof you don't need to apply the cauchy definition of a limit but you have to use the intermediate value theorem, which does need to use that definition.

So technically there is no answer.

Let[math]x \,\in\, \mathbf R \,\setminus\, 2\,\pi\,\mathbf Z[/math].

[eqn]1 \,-\, \cos\,x \,=\, \frac{x^2}{2} \,+\, \underset{x \,\rightarrow\, 0}{o} \left( x^2 \right)[/eqn]

Then

[eqn]\frac{x^2}{1 \,-\, \cos\,x} \,=\, \frac{x^2}{\frac{x^2}{2} \,+\, \underset{x \,\rightarrow\, 0}{o} \left( x^2 \right)} \,=\, \frac{2}{1 \,+\, \underset{x \,\rightarrow\, 0}{o} \left( 1 \right)} \,\xrightarrow[x \,\rightarrow\, 0]{}\, 2[/eqn]

Therefore [math]\lim_{x \,\rightarrow\, 0} \frac{x^2}{1 \,-\, \cos\,x} \,=\, 2[/math].

L'Hôpital's rule is a fucking meme. Never use it.

>>8514445
If x=0, what's the ratio look like?
Now if x<0? Or x>0? Maybe you're thrown by the trig

>>8514881
Did you mean
>[math]\color{#b5bd68}{1 - \cos x = \dfrac{x^2}{2}+\mathcal{O}(x^3)}[/math]
?

You need differentiation to justify this you chucklefuck. That's exactly what OP is trying to avoid by not using L'H.

>L'Hôpital's rule is a fucking meme
Stupid freshmen.

File: gorach.jpg (74KB, 800x600px) Image search: [Google]

74KB, 800x600px

>>8514445
>>8514457
>>8514463
>>8514881
You're all retarded.
lim x->0 ((x^(2))/((1)-(cos(x)))
ctrl c +ctrl v into wolfram alpha.
Again, stop mental masturbating and stop listening to ivy fags.

>>8514777
Doesn't not you need the function f(x) and g(x) on a certain interval [a,b] satisfy the Cauchy theorem so that you can use l'hospital?

>>8514445
kek, my calc 1 teacher did the same thing but he never let us use l'hopital's because he said it's too easy

that one in particular looks like a trig identity waiting to happen, try multiplying the numerator and denominator by 1+cosx

>>8514926
WA uses l'hospital

>>8514922
>Calls someone a freshman chucklefuck
>Fucks up a one-line [math]\LaTeX[/math] formula, completely misunderstands OP's problem and would use l'Hôpital's meme for a limit this simple

>>8514926
Wolfcuck Alpha uses l'Hôpital's rule for all limit calculations you fucking chemfag.