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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 $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$
Sep
22
asked Solving $\lim _{x\to 1}\left(\frac{1-\sqrt[3]{4-3x}}{x-1}\right)$
Sep
22
accepted Solving $\lim_{x \to 4}\frac{3-\sqrt{5+x}}{1-\sqrt{5-x}}$ without L'Hopital.