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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
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
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
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
7
comment For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of this system?
Thanks, that's right (although $b$ yields $-5$).
Sep
7
comment For what values of $p$ and $b$ is the vector $(b,8,b+7)$ a solution of this system?
@Adriano: Huh, it seems it should be $b = -5$. The exercise's answer seems to be wrong haha.
Sep
1
comment Is $A$ invertible if $ABAB^2 = I$?
Only of the same size?
Aug
21
comment Factoring $x^3-8$ by grouping
@Winther that works - but how did you figure that? Just made up?
Aug
21
comment Factoring $x^3-8$ by grouping
Sorry, how did you realize that $x^3 - 8 = (x^3 + 2x^2 + 4x) + (-2x^2 - 4x - 8)$?
Jul
5
comment Calculating an orthonormal base given another base.
Aha, this worked! I'll keep on testing it. Do you know why my particular approach didn't work?
Jul
5
comment Calculating an orthonormal base given another base.
By "second vector" you mean $(a,b,c)$?
Jul
3
comment Finding the smallest $x$ given a set of congruence conditions.
Did you mean $123$ instead of $133$?
Jun
22
comment How can a subspace have a lower dimension than its parent space?
Yes, that's exactly it! Thank you.
Jun
22
comment How can a subspace have a lower dimension than its parent space?
I'm sorry, I made a mistake with my example - the real core question is, how can a subspace have a lower dimension than its parent space? (according to the above theorem) - that's why I made up that example.
Jun
17
comment Understanding the orthogonal complement of a subspace.
Ahhh it all makes sense. Yes, thank you.
Jun
17
comment Understanding the orthogonal complement of a subspace.
And the orthogonal complement of a line in 3D is another line, too? Is a plane ever the orthogonal complement of something?
Jun
17
comment About an orthogonal complement theorem
@JulianP: You're right! I guess that's the problem?
Jun
16
comment Coordinate vector of a subspace of $\mathbb{M}_{2,2}(\mathbb{R})$
@user84413: I'm sorry, I didn't know that. I have added the basis. $\left\{ \left ( \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right ) , \left ( \begin{matrix} 0 & 1 \\ 1 & 1 \end{matrix} \right )\right \}$
Jun
16
comment Coordinate vector of a subspace of $\mathbb{M}_{2,2}(\mathbb{R})$
@user156384: No. It may have been an error I guess. Just curious if perhaps there were multiple coordinate vectors (can there be?)