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Finding a Limit Without L’hôpital Rule or Taylor Expansion

Finding a Limit Without L’hôpital Rule or Taylor Expansion

Finding a Limit Without L’hôpital Rule or Taylor Expansion


Posted: May 19, 2022 in Mathematics
Tags: exams, limit, math, solution

Finding a Limit Without L’hôpital Rule or Taylor Expansion

Saw this limit when I was watching a Korean drama, Melancholia. The drama is about a romance love (and math) story between a math prodigy and his math teacher. In the drama, the school principal and one of the math teachers had been helping out the daughter of a politician cheating in the math tests and exams so she can always be the top of the class. In one of the exams, the main actress (one of the math teachers) has modified one of the questions due to the error or typo in one question. She didn’t notify her colleagues she modified one of the questions and of course her colleague later gave the full solution (the solution that is based on exam that contains the error) to the politician’s daughter to memorize before taking the exam. Later on, the main actress and other teachers suspected the politician’s daughter was cheating. To avoid the investigation, the principal and the math teacher who leaked the solution to the politician’s daughter came up with the idea to let the whole class to re-take the exam. By using an excuse that in one of the exam problems requires the knowledge that is beyond the syllabus. Here is the limit,

\displaystyle \lim_{x\to0} \frac{\tan x - x}{x^3}

Of course, the main actor, the math prodigy, told the teachers that he did not use L’hôpital rule to find the limit. Since the movie is in Korean and the subtitle is in simplified Chinese (殘體字). I don’t quite understand what method he used. But I used his idea replacing x with 2x and solve with my own method,

\displaystyle L = \lim_{x\to0} \frac{\tan x - x}{x^3} = \lim_{2x\to0} \frac{\tan{2x}-2x}{(2x)^3} = \frac{1}{4}\lim_{x\to0}\frac{\frac{1}{2}\tan{2x}-x}{x^3}

hence,

\displaystyle 4L = \lim_{x\to0} \frac{\frac{1}{2}\tan{2x}-x}{x^3}

and so we subtract 4L by L and get

\displaystyle 3L = \lim_{x\to0} \frac{\frac{1}{2}\tan{2x}-\tan x}{x^3} = \lim_{x\to0} \frac{1}{x^3}\left(\frac{\tan x}{1-\tan^2 x}-\tan x\right)

\displaystyle = \frac{\tan x}{x^3} \left(\frac{1-1+\tan^2 x}{1-\tan^2 x}\right) = \lim_{x\to0} \left(\frac{\tan x}{x}\right)^3 \frac{1}{1-\tan^2 x}

\displaystyle = \lim_{x\to0} \left(\frac{\sin x}{x}\right)^3 \cdot \frac{1}{\cos^3 x} \cdot \frac{1}{1-\tan^2 x} = 1

\displaystyle \therefore L = \frac{1}{3}

I also saw a legendary IMO question 6 problem on the later episode of the drama, I will leave it for my next blog so stay tuned.


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