Question

limits as x goes to negative infinity

score 391 · Answer 1 · 2026-03-25 01:15:18

391

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limits as x goes to negative infinity

Published on 25 Mar 2026 - 1:15

score 210 · Answer 2 · 2026-03-22 23:35:13

210

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Sum of the series $\csc^{-1} \sqrt{10}+ \csc^{-1} \sqrt{50}+\csc^{-1}\sqrt{170}...$

Published on 22 Mar 2026 - 23:35

score 56 · Answer 3 · 2018-02-11 21:16:48

56

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How to find the values that $x$ can take to be a real number?

Published on 11 Feb 2018 - 21:16

score 462 · Answer 4 · 2026-03-25 01:22:34

462

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Compute $\sqrt{5}$ using Newton's method and regula falsi method

Published on 25 Mar 2026 - 1:22

score 83 · Answer 5 · 2026-03-14 03:54:37

83

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Proving $ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right) $

Published on 14 Mar 2026 - 3:54

score 144 · Answer 6 · 2018-02-13 13:24:43

144

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Show that if $x, y, \sqrt{x} + \sqrt{y}∈ \mathbb Q$ then $\sqrt{x}, \sqrt{y} ∈ \mathbb Q$

Published on 13 Feb 2018 - 13:24

score 487 · Answer 7 · 2018-02-15 01:19:12

487

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Determining point of discontinuity of a rational function with a root

Published on 15 Feb 2018 - 1:19

score 60 · Answer 8 · 2018-02-15 14:52:42

60

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Question about $\lim_{n\rightarrow\infty} (\sqrt{n} - \sqrt{n - n^c})$ and $\lim_{n\rightarrow\infty} (\sqrt[3]{n + n^c} - \sqrt[3]{n})$

Published on 15 Feb 2018 - 14:52

score 107 · Answer 9 · 2026-03-20 05:03:32

107

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maximum value of $a^2b$ condition is given

Published on 20 Mar 2026 - 5:03

score 196 · Answer 10 · 2026-03-20 02:57:02

196

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Proof that $\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{1+2(\ldots)}}}} = 1+\sqrt{2}$

Published on 20 Mar 2026 - 2:57

score 101 · Answer 11 · 2018-02-18 14:31:49

101

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What does it mean to be equal?

Published on 18 Feb 2018 - 14:31

score 868 · Answer 12 · 2018-02-18 22:05:29

868

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$\frac{\sqrt{26} - \sqrt{13} - \sqrt{2}+1}{\sqrt{13} -1 }= ? $

Published on 18 Feb 2018 - 22:05

score 162 · Answer 13 · 2018-02-18 23:01:47

162

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$\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}$

Published on 18 Feb 2018 - 23:01

score 78 · Answer 14 · 2026-03-25 01:18:44

78

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If $a \ge 0$ and $b \ge 0$, then prove that $a \le b \Leftrightarrow a^2 \le b^2 \Leftrightarrow \sqrt{a} \le \sqrt{b}$

Published on 25 Mar 2026 - 1:18

score 488 · Answer 15 · 2026-03-19 23:24:03

488

Views

Solving Radical Inequalities

Published on 19 Mar 2026 - 23:24

List Question

limits as x goes to negative infinity

Sum of the series $\csc^{-1} \sqrt{10}+ \csc^{-1} \sqrt{50}+\csc^{-1}\sqrt{170}...$

How to find the values that $x$ can take to be a real number?

Compute $\sqrt{5}$ using Newton's method and regula falsi method

Proving $ \frac {(a+b)^2 }{\sqrt{(a+b)^2 +1}} < 2 \left( \frac {a^2 }{\sqrt{ a^2 +1}} + \frac{b^2 } {\sqrt{ b^2 +1 }} \right) $

Show that if $x, y, \sqrt{x} + \sqrt{y}∈ \mathbb Q$ then $\sqrt{x}, \sqrt{y} ∈ \mathbb Q$

Determining point of discontinuity of a rational function with a root

Question about $\lim_{n\rightarrow\infty} (\sqrt{n} - \sqrt{n - n^c})$ and $\lim_{n\rightarrow\infty} (\sqrt[3]{n + n^c} - \sqrt[3]{n})$

maximum value of $a^2b$ condition is given

Proof that $\sqrt{1+2\sqrt{1+2\sqrt{1+2\sqrt{1+2(\ldots)}}}} = 1+\sqrt{2}$

What does it mean to be equal?

$\frac{\sqrt{26} - \sqrt{13} - \sqrt{2}+1}{\sqrt{13} -1 }= ? $

$\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}$

If $a \ge 0$ and $b \ge 0$, then prove that $a \le b \Leftrightarrow a^2 \le b^2 \Leftrightarrow \sqrt{a} \le \sqrt{b}$

Solving Radical Inequalities

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