# Calculate the limit $\displaystyle\lim_{x\to0}\frac {(\tan(x)-x)^2} {(\ln(x+1)-x)^3}$

Calculate: $$\displaystyle\lim_{x\to0}\frac {(\tan(x)-x)^2} {(\ln(x+1)-x)^3}$$

So if we expand Taylor polynomials we get:

$$\frac {(x+{x^3\over3}+o(x^3)-x)^2}{(x-{x^2\over2}+o(x^2)-x)^3}=\frac {({x^3\over3}+o(x^3))^2}{(-{x^2\over2}+o(x^2))^3}$$

How do I continue from here ?

Another related question, what is the best way to memorize the taylor polynomial of all of the common elementary functions ?

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You are missing a minus sign in the denominator. Numerator and denominator have the same leading order, $x^6$. Divide both by that. – Daniel Fischer Feb 26 '14 at 12:24
Just drop all the ${} + o(...)$ since they are too small to be considered; and proceed from there. – user49685 Feb 26 '14 at 12:25
@user49685 Is it formal to just drop them because they're small ? – GinKin Feb 26 '14 at 12:29
@GinKin: Yes, you can drop them as long as the new expression obtained by dropping them, doesn't belong to any indeterminate forms. – user49685 Feb 26 '14 at 17:21

I think it should be$$\frac {({x^3\over3}+o(x^5))^2}{(-{x^2\over2}+o(x^3))^3}$$ and so $$\frac {x^6({1\over3}+o(x^2))^2}{x^6(-{1\over2}+o(x))^3}\to \frac{({1\over3})^2}{({-1\over2})^3} \ \ \text{as }\ \ x\to 0$$

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you are completely right, thank you – Sameh Shenawy Feb 26 '14 at 14:23
Isn't it necessary to include some explanation when removing the little os ? – GinKin Feb 26 '14 at 16:12

This is based on two standard limits $$\lim_{x \to 0}\frac{\tan x - x}{x^{3}} =\lim_{x \to 0}\frac{\sec^{2}x - 1}{3x^{2}}\text{ (by LHR)} = \frac{1}{3}\lim_{x \to 0}\left(\frac{\tan x}{x}\right)^{2} = \frac{1}{3}$$ and $$\lim_{x \to 0}\frac{\log(1 + x) - x}{x^{2}} = \lim_{x \to 0}\dfrac{\dfrac{1}{1 + x} - 1}{2x}\text{ (by LHR)} = -\frac{1}{2}$$ Clearly we now have \begin{aligned}L &= \lim_{x \to 0}\frac{(\tan x - x)^{2}}{\{\log(1 + x) - x\}^{3}}\\ &= \lim_{x \to 0}\dfrac{\left(\dfrac{\tan x - x}{x^{3}}\right)^{2}}{\left(\dfrac{\log(1 + x) - x}{x^{2}}\right)^{3}}\text{ (dividing Nr and Dr by }x^{6})\\ &= \dfrac{(1/3)^{2}}{(-1/2)^{3}}= -\frac{8}{9}\end{aligned}

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How can you substitute these two limits and why the curly braces ? – GinKin Feb 26 '14 at 15:45
@GinKin: About substitution you know the rule $\lim_{x \to a}f(x)/g(x) = \lim_{x \to a}f(x)/\lim_{x \to a}g(x)$ provided both limits exist and denominator limit is non-zero. And the curly braces are for clarity. I would change them to normal parentheses. Don't worry for that. – Paramanand Singh Feb 26 '14 at 15:49
I still don't see how can you substitute them so $\lim (\tan x - x)^{2}=\lim\dfrac{\tan x - x}{x^{3}}$ and the same for the other one... – GinKin Feb 26 '14 at 15:57
@GinKin: I have divided numerator and denominator by $x^{6}$ and then made substitution. – Paramanand Singh Feb 26 '14 at 15:59
Well in this case the division by $x^{6}$ led to some pretty famous limits in both numerator and denominator and that's why it worked. – Paramanand Singh Feb 26 '14 at 16:23