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 Oct 24 accepted The normal equation of the plane that contains the line $(1,1,1) + t(-2,0,3)$ Oct 24 accepted Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane? Oct 24 comment Why isn't the orthogonal vector to a direction vector of a plane not necessarily perpendicular to such plane? Would the same apply parallel vectors? I mean, if I find a vector that is parallel to that line's direction vector, would it also be parallel to the plane? 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 $3$?