$$\lim_{x\to 0 } \frac {\tan^2x - \arctan^2x}{x^4}$$

I am having trouble calculating the above limit. You could apply L'Hopital's rule 5 times battling with various terms but there must be a simpler way to express tanx / arctanx in another form.

Anyone care to help?

  • 4
    $\begingroup$ If you want an easier time applying l'Hopital (or anything else, really), I suggest rewriting it into $$\left(\lim_{x \to 0}\frac{\tan x + \arctan x}{x}\right)\cdot \left(\lim_{x \to 0}\frac{\tan x - \arctan x}{x^3}\right)$$You still need to use it once of the first limit and three times on the second one, but I imagine the battling with various terms becomes a lot easier. $\endgroup$ – Arthur Mar 8 '16 at 19:50
  • 2
    $\begingroup$ We only need the first two terms of the MacLaurin series for $\tan x$, and for $\arctan x$. $\endgroup$ – André Nicolas Mar 8 '16 at 19:53
  • $\begingroup$ @Arthur, the derivative of the numerator in your second limit, $\sec^2x-{1\over x^2+1}$, can be rewritten as $\tan^2x+{x^2\over x^2+1}$, which makes for considerable simplification. $\endgroup$ – Barry Cipra Mar 8 '16 at 20:13

Note that this limit cannot be solved without the use of Taylor series or L'Hospital's Rule. A solution via Taylor series is already presented by Olivier Oloa and I present one with L'Hospital's Rule.

Note that if one wishes to apply L'Hospital's Rule to solve any limit one should try to apply it only when needed. Thus we first make use of algebraic simplification and standard limits and then use L'Hospital's Rule.

We have \begin{align} L &= \lim_{x \to 0}\frac{\tan^{2}x - \arctan^{2}x}{x^{4}}\notag\\ &= \lim_{x \to 0}\frac{\tan x + \arctan x}{x}\cdot\frac{\tan x - \arctan x}{x^{3}}\notag\\ &= \lim_{x \to 0}\left(\frac{\tan x}{x} + \frac{\arctan x}{x}\right)\cdot\frac{\tan x - \arctan x}{x^{3}}\notag\\ &= \lim_{x \to 0}\left(1 + 1\right)\cdot\frac{\tan x - \arctan x}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\tan x - x + x - \arctan x}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}} + \frac{x - \arctan x}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}} + 2\lim_{x \to 0}\frac{x - \arctan x}{x^{3}}\tag{1}\\ &= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}} + 2\lim_{x \to 0}\frac{x - \arctan x}{\arctan^{3}x}\cdot\frac{\arctan^{3}x}{x^{3}}\notag\\ &= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}} + 2\lim_{x \to 0}\frac{x - \arctan x}{\arctan^{3}x}\cdot 1\notag\\ &= 2\lim_{x \to 0}\frac{\tan x - x}{x^{3}} + 2\lim_{t \to 0}\frac{\tan t - t}{t^{3}}\text{ (putting }t = \arctan x)\notag\\ &= 4\lim_{x \to 0}\frac{\tan x - x}{x^{3}}\tag{2}\\ &= 4\lim_{x \to 0}\frac{\sec^{2}x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{4}{3}\lim_{x \to 0}\frac{\tan^{2}x}{x^{2}}\notag\\ &= \frac{4}{3}\notag \end{align} Steps from $(1)$ to $(2)$ are based on the assumption that the limit $\lim_{x \to 0}\dfrac{\tan x - x}{x^{3}}$ exists and this assumption is justified in the steps after $(2)$ via L'Hospital's Rule.

Note that L'Hospital's Rule has been used only once (compared to what OP mentions in his post) and the standard limit $$\lim_{x \to 0}\frac{\tan x}{x} = 1$$ or equivalently $$\lim_{x \to 0}\frac{\arctan x}{x} = 1$$ has been used in the above solution.

  • $\begingroup$ After (2) we can use taylor expansion,too. $\endgroup$ – Takahiro Waki Mar 9 '16 at 14:11

One may use some standard Taylor expansions as $x \to 0$.

We have $$ \tan x =x+\frac{x^3}3+O(x^5) $$ giving $$ \tan^2 x =x^2+\frac{2x^4}3+O(x^6) $$ and $$ \arctan x =x-\frac{x^3}3+O(x^5) $$ giving $$ \arctan^2 x =x^2-\frac{2x^4}3+O(x^6). $$ Then

$$ \frac{\tan^2 x-\arctan^2 x}{x^4}=\frac43+O(x^2) $$


$$ \lim_{x \to 0}\frac{\tan^2 x-\arctan^2 x}{x^4}=\frac43. $$


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