Zol Tun Kul
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 Sep 28 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ I used the two things you showed me and progressed to $$\frac{-6-x}{\sqrt{4-\frac{6}{x^2}}+\sqrt{4+\frac{1}{x}}}$$. While it does seem a bit better it is still not quite clear to me what to do with that. Sep 28 comment Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Does the first identity have a name I can search for? Sep 28 asked Calculating $\lim_{x\to-\infty}\left(\sqrt{4x^2-6}-\sqrt{4x^2+x}\right)$ Sep 27 accepted Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ Sep 27 comment Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ What is $o(x)$? Sep 27 comment Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I don't really see how does multiplying by $\frac{(1+\cos(x))}{(1+\cos(x))}$ help me. Sep 27 comment Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ Hm. I do know that $\frac{\sin x}{x} = 1$, but I am not quite sure why does that imply that $\frac{1-\cos x}{x^2} = \frac{1}{2}$. Sep 27 asked Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ Sep 27 accepted Evalutating $\lim_{x\to0}\frac{1-\cos x}{x^2}$ Sep 27 comment Evalutating $\lim_{x\to0}\frac{1-\cos x}{x^2}$ That's definitely it. Thank you. Sep 27 asked Evalutating $\lim_{x\to0}\frac{1-\cos x}{x^2}$ Sep 27 accepted Calculating $\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$ Sep 26 comment Calculating $\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$ Why is $1-\cos x = 2 \sin^2 \frac{1}{2}x$? Is it some sort of theorem that permits that equivalence? Sep 26 asked Calculating $\lim_{x\to 0}\frac{\sec (x)-1}{x^2\sec(x)}$ Sep 26 accepted Why is $|x| = -x$ in $\lim_{x\to-\infty}\frac{3-x}{\sqrt{2x^2-1}}$? Sep 26 comment Why is $|x| = -x$ in $\lim_{x\to-\infty}\frac{3-x}{\sqrt{2x^2-1}}$? @Aleksandar Eh, I copied it from the lecture but I don't quite remember the reasoning behind it. Sep 26 asked Why is $|x| = -x$ in $\lim_{x\to-\infty}\frac{3-x}{\sqrt{2x^2-1}}$? Sep 25 accepted Solving $\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$ without L'Hopital Sep 24 asked Solving $\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$ without L'Hopital Sep 22 accepted Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$