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Oct
24
asked Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane?
Oct
24
asked The normal equation of the plane that contains the line $(1,1,1) + t(-2,0,3)$
Oct
14
awarded  Popular Question
Oct
3
accepted Deriving $\frac{8}{\sqrt{x-2}}$
Oct
2
awarded  Notable Question
Oct
2
asked Deriving $\frac{8}{\sqrt{x-2}}$
Oct
1
comment Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
By the way, if $h \not = 0$, how come the denominator is $(x-1)^2$? I had said that "$h$ is practically $0$" but I'm not that convinced anymore.
Oct
1
accepted Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
Oct
1
asked Why $(-1 \cdot h) = -1$ when $h$ approaches $0$?
Sep
29
awarded  Popular Question
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@AndréNicolas I see now. Thanks. By the way, you picked $\frac{1}{4}$ because it made arithmetic easier - but how can I determine the "largest acceptable value" that can replace $\frac{1}{4}$? (out of curiosity)
Sep
20
accepted Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@EWHLee I see. I wonder if $\frac{1}{4x-9}$ should have been $\frac{1}{|4x-9|}$ instead... Wait, no, that doesn't work either. Goddammit. What should I have done instead?
Sep
20
comment Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
@AndréNicolas Ahhhh yes, that's right. By the way, isn't $|(x-3)| < \delta < \frac{9}{4}$ enough to keep $|4x-9|$ away from $0$ since $\frac{9}{4}$ is the solution to $4x-9$?
Sep
20
asked Proving that $\lim_{x\to3}\frac{x}{4x-9}=1$
Sep
20
asked Why is $|(x-2)| < \delta \le 1$ true when proving $\lim_{x\to2}(3x^2-x)=10$?
Sep
19
comment Proving $\lim_{x\to1}(x^2+3)=4$
Hm I'm not very sure how can $|x-1| < \delta$ lead to $|x+1|\le 3$. If you increase $|x-1|$ by $2$ to reach $|x+1|$, shouldn't it be $|x+1| \le 2$ rather than $3$?
Sep
19
accepted Proving $\lim_{x\to1}(x^2+3)=4$
Sep
19
comment Why does $|x-1|^2+3|x-1| < \epsilon \implies |x-1|^2 < \frac{\epsilon}{2} \ \ \land \ \ 3|x-1| < \frac{\epsilon}{2}$?
@Oleg567: What if $a = b = \epsilon/2$?
Sep
19
asked Why does $|x-1|^2+3|x-1| < \epsilon \implies |x-1|^2 < \frac{\epsilon}{2} \ \ \land \ \ 3|x-1| < \frac{\epsilon}{2}$?