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I had trouble evaluating this limit and found this solution from another post of Stack Exchange. The point of this solution was trying to find the limit without using Taylor's series or L'Hospital's rule.

I just need help understanding certain steps:

  1. How did he go from step 3 to step 4.

  2. How did he go from step 4 to step 5.

His solution: Description

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    $\begingroup$ Apologies, I forgot to add the description so the link disappeared. $\endgroup$ – A.AK Feb 11 '17 at 18:39
  • $\begingroup$ While the solution presented above is smart, it only shows that "if the limit exists then it must be $1/3$". This is different from showing that "the limit exists and is equal to $1/3$". $\endgroup$ – Paramanand Singh Feb 12 '17 at 7:51
  • $\begingroup$ How would you prove that it exists? $\endgroup$ – A.AK Feb 12 '17 at 8:14
  • $\begingroup$ There is no easy way to show that it exists unless you use L'Hospital's Rule or Taylor's theorem. $\endgroup$ – Paramanand Singh Feb 12 '17 at 8:17
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If we have:

$$4L=\lim_{x\to 0} \frac{\frac12\tan 2x-x}{x^3} \qquad L=\lim_{x\to 0} \frac{\tan x-x}{x^3}$$

substracting we would get

$$4L-L=3L=\lim_{x\to 0} \frac{\frac12\tan 2x-\tan x}{x^3}$$

Now, use that $\tan(2x)=\frac{2\tan x}{1-\tan^2 x}$ to go on:

\begin{equation*} \begin{split} 3L&=\lim_{x\to 0} \frac{\frac{\tan x}{1-\tan^2x}-\tan x}{x^3}=\lim_{x\to 0} \frac{\tan x}{x} \cdot \frac{\frac{1}{1-\tan^2 x}-1}{x^2}=\lim_{x\to 0} \frac{\tan x}{x}\cdot \frac{\tan^2 x}{x^2(1-\tan^2 x)}=\\=&\lim_{x\to 0} \frac{\tan x}{x}\cdot \frac{\tan^2 x}{x^2}\cdot\frac{1}{1-\tan^2 x}=\lim_{x\to 0} \frac{\tan^3 x}{x^3} \cdot \lim_{x\to 0} \frac{1}{1-\tan x^2}\end{split} \end{equation*}

Now, as $\lim_{x\to 0} \frac{1}{1-\tan^2 x}=1$, (because $\tan x\to 0$ when $x\to 0$) and $$\lim_{x\to 0} \frac{\tan^3 x}{x^3}=\left(\lim_{x\to 0} \frac{\tan x}{x}\right)^3=1^3=1$$ we get that $3L=1$.

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  • $\begingroup$ Could you please explain step 5 to step 6 as well? $\endgroup$ – A.AK Feb 11 '17 at 18:54
  • $\begingroup$ You are welcome~ $\endgroup$ – A. Salguero-Alarcón Feb 11 '17 at 19:11

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