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I have a belief that it is possible to evaluate any limit without using L'Hospital rule or series expansions. However the limit $$\lim\limits_{x\to 0} \frac{x^2\cos x - 6 \ln(1+x^2) +5x^2}{(e^{\sqrt[4]{1+4x^3+8x^4}}-e)\arcsin(x)}$$ stumps me.

I thought at the beginning that is it nothing but dividing top and bottom by $x^2$ and then using the limits: $$\lim\limits_{x\to 0} \frac{\ln(1+x^2)}{x^2} = 1, \quad \quad \lim \limits_{x\to 0} \frac{\arcsin x}{x} = 1$$ And then I found $$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x} = 0$$

Which made the limit $\frac{0}{0}$. I have no clue how to split the limit into a product or a sum of existent limits. Is it possible to solve this without L'Hosptial? Is there is possible usage of the squeeze theorem or a geometric interpretation to compute this limit? The answer according to WolframAlpha is $\frac{5}{2e}$ and it is clear how to find the $\frac{1}{e}$ but not the $\frac{5}{2}$.

MathFail
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geoalg
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    What is the source of this problem? If from a book or class, which class? Also, what is the exact wording of the original problem? For example, does the original problem specifically require that L'Hopital's Rule not be used. This strikes me as unusual, for such a complicated problem. ...see my next comment – user2661923 Aug 02 '22 at 20:53
  • if you're dividing both sides by $x^2$, then in the denominator you should have just 1 $x$ on the arcsine, and 1 $x$ on the exponential. – Angelica Aug 02 '22 at 20:54
  • You'll need to divide the numerator and denominator by $x^4$, as they both have fourth-order zeros. – eyeballfrog Aug 02 '22 at 20:54
  • Assuming that this problem is from a book or class, I advise metacheating, explained as follows: First, see the Edit-Tools section of this article. The point here is not that you haven't shown work. Instead, for such a complicated problem, the idea is that the problem composer must have provided highly specialized tools that can be used to attack such a problem. So, the meta-cheating first question is: What are those tools? – user2661923 Aug 02 '22 at 20:56
  • @user2661923 some people just want to compute limits without lhopital or series. I am one of those people, and I am not alone. – Ninad Munshi Aug 02 '22 at 20:59
  • @user2661923 I fit in the category that Ninad just described. As for the source of the problem, this problem was raised as a counterexample to my belief that any limit can be computed without using the L'Hopital rule or series expansions. I am not sure if it comes from a book though. – geoalg Aug 02 '22 at 21:01
  • So, you are artificially imposing the constraint against L'Hopital's rule yourself? This is an interesting challenge. The difficulty is that in such a scenario (below the level of graduate school), it can be frustrating to try to attack such a challenge without training. My objection here is that such frustration does not help Math education. – user2661923 Aug 02 '22 at 21:02
  • @user2661923 you're not wrong, but you're also missing the point. It's kind of the same argument I hear against integration bee style integrals all the time. Yes it's true that it would be frustrating educationally without training - but we aren't here without training. We just want a challenge that's quick but requires thought like a crossword puzzle. – Ninad Munshi Aug 02 '22 at 21:04

2 Answers2

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With the hint of eyeballfrog, I was able to arrive at the following solution.

We write the nuemreator as $x^2(\cos x - 1) - 6(\ln(1+x^2)-x^2)$ then dividing top and bottom by $x^4$ the limit becomes $$\lim\limits_{x\to 0} \frac{\frac{\cos x - 1}{x^2}-\frac{6}{x^4}(\ln(1+x^2)-x^2)}{\frac{\arcsin x}{x} \cdot \frac{e^{\sqrt{1+4x^3+8x^4}-1}-1}{x^3}} $$

The numerator can be computed using the limits $\lim\limits_{x\to 0}\frac{\cos x -1}{x^2} = -\frac{1}{2}$ and $\lim\limits_{x\to 0} \frac{\ln(1+x^2)-x^2}{x^4} = \frac{-1}{2}$ where the latter is proved in Are all limits solvable without L'Hôpital Rule or Series Expansion

For the bottom, we use the limits in the original post except now it is more divided by $x^3$ $$\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3} = e\lim\limits_{x\to 0} \frac{e^{\sqrt[4]{1+4x^3+8x^4}-1}-1}{\sqrt[4]{1+4x^3+8x^4}-1} \cdot \frac{\sqrt[4]{1+4x^3+8x^4}-1}{(1+4x^3+8x^4)-1} \cdot \frac{4x^3+8x^4}{x^3} = e \cdot \frac{1}{4} \cdot 4$$ Combining all of this we arrive at the desired result $\frac{5}{2e}$

geoalg
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The numerator is $\sim x^2(\cos(x)-1)+3x^4\sim \frac{5}{2}x^4$

The denominator is $\sim (e^{\sqrt[4]{1+4x^3+8x^4}}-e)x$

So you need to evaluate the limit for

$$\frac{e^{\sqrt[4]{1+4x^3+8x^4}}-e}{x^3}$$

You did $x^2$ instead of $x^3$, so you got $0$, and you can show this term $$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e\cdot x^3$$

Series expansion $\sqrt[4]{1+y}=1+\frac{1}4y+O(y^2)$, substitute $y=4x^3+8x^4$

$$e^{\sqrt[4]{1+4x^3+8x^4}}-e\sim e^{1+x^3+2x^4}-e=e\cdot(e^{x^3+2x^4}-1)\sim e\cdot x^3$$

Final answer:

$$L=\frac{5}{2e}$$

MathFail
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