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find the limit.

$$\lim_{x\to0^{+}}\dfrac{\tan{(\tan{x})}-\tan{(\sin{x})}}{\tan{x}-\sin{x}}$$

my try: $$\tan{x}=x+\dfrac{1}{3}x^3+o(x^3),\sin{x}=x-\dfrac{1}{6}x^3+o(x^3)$$

so $$\tan{(\tan{x})}=\tan{x}+\dfrac{1}{3}(\tan{x})^3+o(\tan^3{x})=x+\dfrac{1}{3}x^3+\dfrac{1}{3}(x+\dfrac{1}{3}x^3)^3+o(x^3)$$ $$\tan{(\sin{x})}=\sin{x}+\dfrac{1}{3}(\sin{x})^3+o(\sin^3{x})=x-\dfrac{1}{6}x^3+\dfrac{1}{3}(x-\dfrac{1}{6}x^3)^3+o(x^3)$$ so $$\tan{(\tan{x})}-\tan{(\sin{x})}=\dfrac{1}{2}x^3+o(x^3)$$ $$\tan{x}-\sin{x}=\dfrac{1}{2}x^3+o(x^3)$$

Have other metods? Thank you

4 Answers4

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I think this identity is useful here:

$$\tan(a-b)=\frac{\tan(a)-\tan(b)}{1+\tan(a)\tan(b)}$$ and the fact that $\tan(a)\sim a$ while $a\sim 0$.

Guy Fsone
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Mikasa
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  • Perfect identity for the task! +1 – amWhy Sep 12 '13 at 00:06
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    Unless I am missing something, to apply this, we would say $$ \frac{\tan(a)-\tan(b)}{a-b}=\frac{\tan(a-b)}{a-b}(1+\tan(a)\tan(b)) $$ but then we need to use the Mean Value Theorem (or L'Hospital) on the right. It would seem more efficient simply to use the MVT directly. – robjohn Sep 12 '13 at 09:19
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Yours is easily the simplest $direct$ method.

My first idea for the limit is that that really looks like a difference quotient for the derivative of $\tan$. Distilling the core idea, I make the following conjecture:

Let $f(x)$ be continuously differentiable at $a$. Then, $$ \lim_{\substack{(x,y) \to (a,a) \\ x \neq y}} \frac{f(x) - f(y)}{x-y} = f'(a)$$

which lets us immediately see the limit should be

$$\tan'(0) = \sec^2(0) = 1$$

4

Since all the functions involved are continuosly differentiable we can apply the Mean Value Theorem to get $$ \frac{\tan(\tan x)-\tan(\sin x)}{\tan x-\sin x}=\frac1{\cos^2 z} $$ for some $z(x)$ with $\sin x\leq z(x)\leq\tan x$. By the Squeeze Rule, $z\to0$ when $x\to0$, and so the limit is $1/\cos^20=1$.

Martin Argerami
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1

Factor a $\tan(x)$ from the denominator: $$\frac{\tan\tan(x) - \tan\sin(x)}{\tan(x)-\sin(x)} = \frac{\frac{\tan\tan(x)}{\tan(x)} - \cos(x)\frac{\tan\sin(x)}{\sin(x)}}{1-\cos(x)}.$$ The quotients in the numerator go to $1$, so the numerator goes to $1-\cos(x)$. This suggests cancellation.

After two derivatives, the denominator of this quotient will go to $1$, so one strategy is L'Hopital's rule. (I don't have time to compute it but maybe I'll have time later today.)

Neal
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