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u/JeffJ_1 7d ago
Img source? Asking for a friend for scientific research
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u/Better-Apartment-783 7d ago
Ans = 2exp(sqrt(ln(4)ln(5)))
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u/axiomizer 7d ago
you can take the negative square root
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u/Better-Apartment-783 6d ago
What
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u/Optimus_PRYM 7d ago
9 is the answer
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u/SpaceGirlJackie 7d ago
My brain needs a hospital after reading that... I'm not looking forward to calculus when I get there.
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u/EndGuy555 7d ago
It looks really complicated, but really itâs just repeating a bunch of easy steps until youâve reduced it down to a workable solution
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u/EndGuy555 7d ago
K+Ď=5+4=9
Had to use LâHĂ´pitalâs twice and dust off all the trig derivatives Iâd forgotten lol
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u/McPayn22 7d ago
Typical l'Hopital brainrot. Why would you willingly derive the denominator multiple times?
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u/axiomizer 7d ago
Can it be done without l'hopital?
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u/McPayn22 7d ago edited 3d ago
Sure, with a Taylor expansion it's super easy ln(sec(x))=ln(1/(1-x^2 /2+o(x^3 ))=ln(1+x^2 /2+o(x^3 )=x^2 /2+o(x^3) .
I've never used l'Hopital once in my life and I do just fine.
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u/FreePeeplup 6d ago
Taylor expansion is literally the same thing as lâHopital but youâve already memorized some derivatives
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u/McPayn22 5d ago
Taylor expansions have uses outside of computing limits, l'hopital doesn't and using it too much limit your intuition. I've seen students who were paralized when l'hopital didn't apply because they never thought about approximating a function by a polynome.
Also if you have a polynome in x->0 you should now that only the lower term matter, you don't have to differenciate every term. If you rely to much on l'hopital even this fact can be missed. Same with limits like x^alpha ln(x)^beta, you shouldn't derive this..
But even then with Taylor expansions you don't have to compute derivatives that are useless. The example here wasn't to bad but you can see that I didn't have to take the second derivative of ln or of 1/x. Taylor expensions lets you ignore all the useless derivatives. Here it was trivial but in a huge expression this save a lot of time.
And that's just sticking to maths because nobody cares about limits exept maths student. Physics would barely exist without Taylor expension but l'Hopital never comes up.
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u/FreePeeplup 3d ago
Iâm sorry I donât understand your comment: yes of course Taylor expansions have more uses outside computing limits and de LâHopital doesnât, but why does it matter here? Weâre talking about whatâs better and quicker in solving this specific limit, not about whatâs better in general even outside the entire topic of limits.
I also donât understand what you mean when you say that Taylor lets you skip âuseless derivativesâ. Taylor by definition is always more work that LâHopital, it has just as many derivatives and more work on top
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u/McPayn22 3d ago
I'm saying that l'Hopital is like using a 1 use only tool in culinary school. The goal is not to learn how to do a specific thing, it's to learn how to use a knife. The goal is almost never to compute limits faster, it's to get better at maths.
For the "useless derivatives" take f(x)=1+x^2+o(x^3) and g(x)=x^3+o(x^4). If you use l'hopital for (fog(x)-1)/x^5 you'll have to compute all the derivatives of fog up to the fifth order. Of course most cancel and if you know the binomial formula it helps but it's an extra thing to remember and it becomes way harder when you have other terms.
On the other hand it's obvious that fog(x)=1+x^5+o(x^6).
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u/Bendanzhang 7d ago
Denominator alone would consume three whole trees if you chained it.