Prove that $$\vert \tan^{-1}x-\tan^{-1}y \vert ≤ \vert x-y\vert \quad \forall x,y \in \mathbb{R}. $$   What I did was to put $x \to x+h$ and $y \to x$ and then take $\lim_{h \to 0}$. This gives $$\lim_{h \to 0} \frac{\vert \tan^{-1}(x+h)-\tan^{-1}x\vert}{\vert h \vert} \le 1$$ which gives $$\frac{d \tan^{-1}x}{dx} = \frac{1}{1+x^2} \le 1.$$ But this proves it only for $x$ and another value $x+h$ very close to it and not $\forall x,y$. Please help.
122086
I have to prove that $|\arctan x - \arctan y|\leq |x-y|$ for all real numbers $x$ and $y$. I know that it has something to do with the mean value theorem but I just don't know how to solve it. All help is appreciated  Thanks in advance :)
34373
Find the limit $$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$  My attempt: Since $$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$ $$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$ So $$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$ Though this method might solve, I think this problem has nicer methods. Thanks.
44502
How to evaluate $\lim_{x\to 0+}\dfrac 1x \Big(\dfrac 1{\tan^{-1}x}-\dfrac 1x\Big)$ ? I used L'Hospital's rule but with no success.
83905
Find $$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)}$$  What do you guys think about this approach, and are there any faster and easier approaches? My approach:   $\frac{\tan(x^3+y^3)(x^2+y^2)}{\sin(x^2+y^2)(x^2+y^2)}$  2.$\frac{\tan(x^3+y^3)}{(x^2+y^2)}$  $\frac{\sin(x^3+y^3)}{\cos(x^3+y^3)\rightarrow 1(x^2+y^2)}$  $\frac{\sin(x^3+y^3)}{(x^2+y^2)}$  Polar: $rcos^3\phi + rsin^3\phi = 0 + 0 = 0$    Thus, $$\lim_{(x,y) \to (0,0)}  \frac{\tan(x^3+y^3)}{\sin(x^2+y^2)} = 0$$
30189
Can you help me to prove that $$|x-y|\leq |x|+|y|$$ I get a proff of this equality, but it's very short and I don't know if it's correct.
77232
I need to use the fact that  $\tan 2x=\sin2x \ /\cos2x$ to prove that: $$\tan 2x=\frac{2\tan x}{1-\tan^2x}$$ I don't know where to start. Please help or hint. Thanks in advance.
114720
I got stuck on this inequality for a day. If $p$ is positive integer, then the problem becomes too easy, but I can't find how we deal with the general case when p can be any positive number. Can someone give me a hint? I really appreciate     Prove that $|x^p - y^p| \le p|x-y|(x^{p-1} + y^{p-1})$ provided that $1 \le p \lt \infty$ and $x, y \ge 0$
81709
I'm trying to prove the inequality  $$|e^{i\theta_1}-e^{i\theta_2}| \leq |\theta_1 - \theta_2|$$  I have tried to use Taylor's formula and got this  $$|e^{i\theta_1}-e^{i\theta_2}| = |(1+i\theta_1 - \frac{\theta_1^2}{2} + \ldots) -(1 +i\theta_2 - \frac{\theta_2^2}{2} + \ldots)| = |i(\theta_1-\theta_2) + \frac{\theta_2^2-\theta_1^2}{2}+\ldots|$$  The first term looks right, but how do I proceed?
88441
I would like to show that show $|t|\leq \pi$ implies $ |e^{it}-1|\geq 2|t|/ \pi$. I tried to proof it geometrically but without success.
51490
I stumbled on the following inequality: For all $n\geq 1,$ $$2\sqrt{n+1}-2\sqrt{n}&lt;\frac{1}{\sqrt{n}}&lt;2\sqrt{n}-2\sqrt{n-1}.$$ However I cannot find the proof of this anywhere. Any ideas how to proceed?  Edit: I posted a follow-up question about generalizations of this inequality here:
74372
Prove $\sin(x)\tan(x) &gt; x^2$ for $x \in ( \,0, \frac{\pi}{2}) \,$  So I did the following: Let $f(x) = \sin(x)\tan(x) - x^2$.  Then of course $f(0) = 0$ and I want to show that for $x \in ( \,0, \frac{\pi}{2}) \,$ $f'(x) &gt; 0$ but I don't really know where to go from here as I can't get anything reasonable done with the derivative.  I was thinking of approximating it using the following inequality $\sin(x) &lt; x &lt; \tan(x)$.   Is that a good direction?
94214
I was solving limits and it said $$\lim_{x\to 0} \left[\frac{x^2}{\tan(x)\sin(x)}\right]=0$$  I tried thinking about it but since I am not really good at math I couldn't proceed.  I know that $\lim_{x\to 0} \left[\frac{\sin(x)}{x}\right] = 0$ and $\lim_{x\to 0} \left[\frac{\tan(x)}{x}\right] = 1$
88959
Prove that $$\frac{x^2}{1+x^4} \leq \frac{1}{2}.$$   First off, we observe that the expression on the LHS is positive for all $x \in \Bbb R,$ and equality is achieved iff $x \in \{-1, 1 \}$. That being said, we start by manipulating the expression as below;  \begin{align} \frac{x^2}{1+x^4} - \frac{1}{2} &amp;\leq 0\\\\ \impliedby\frac{2x^2 - (1+x^4)}{2(1+x^4)} &amp;\leq 0\\\\ \impliedby -\frac{(x^4 - 2x^2 + 1)}{2(1+x^4)}&amp;\leq 0\\\\ \impliedby \frac{x^4 - 2x^2 + 1}{2(1+x^4)} &amp;\geq 0\\\\ \end{align}   We then make the observation that $x^4 - 2x^2 + 1 = (x^2 - 1)^2 \geq 0$ for all $x \in \Bbb R$ and that $1 + x^4 \geq 1$ for all $x \in \Bbb R$, and so the final inequality holds for all $x \in \Bbb R. \\\hspace{92pt}\square$  Is it okay to prove it in this way? I have no solutions so it would be nice if I could have some validation!
139
How do I prove $\sqrt{x + y} \le \sqrt{x} + \sqrt{y}$? for $x, y$ positive?  This should be easy, but I'm not seeing how.  A hint would be appreciated.
91390
how do you prove this inequality? $|\sqrt[n]{x}-\sqrt[n]{y}| \le \sqrt[n]{|x-y|}$ At first glance I thought the triangle inequality would be useful but it's not form what I see. If I raise everything to the power of n I get the following: Possibility 1 (n uneven) : $|x + ... -y| \le |x-y|$ I guess I would have to prove that ... $\le$ 0 but I don't know how to do that. Possibility 2 (n even) : $|x + ... +y| \le |x-y|$ Same as above but now I have to prove that ... $\le$ -y. Am I even on the right track here? Any help would be appreciated. Thanks in advance.
65202
Prove that $\tan20^°\tan40^°\tan60^°\tan80^°=3$    \begin{align} \tan20^°\tan40^°\tan60^°\tan80^°&amp;=\frac{\sin20^°\sin40^°\sin60^°\sin80^°}{\cos20^°\cos40^°\cos60^°\cos80^°} \\ &amp;=\frac{2^4(\sin20^°\sin40^°\sin60^°\sin80^°)^2}{\sin 160^°} \end{align}  I am stuck here.
119149
Prove: $$\frac{1}{\cos^2\alpha}&lt;\frac{\tan\beta-\tan\alpha}{\beta-\alpha}&lt;\frac{1}{\cos^2\beta} \\ \forall 0&lt;\alpha&lt;\beta&lt;\frac\pi2 $$   Well for starters I noticed that this is basically squeeze theorem for derving $\tan x$ since the middle looks like the defintion of deriviation and both left and right sides are the derviation of $\tan x$.  Now to formaly prove it,  $$(\tan\alpha)'&lt;(\tan(\beta-\alpha))'&lt;(\tan\beta)'$$  Then by the next step we get to the given equation and that's it (?).