3h
comment Why do we need to assume continuity in the proof of the chain-rule?
The chain rule does hold assuming only that $f$ is differentiable at $x$. This is as formulated in, e.g., Bartle and Sherbert's Introduction to real Analysis, pg. 162. (I don't see where the proof you give uses the additional assumption.)
10h
comment A limit question with n-th root
$n^2\le n^2+n\le 2n^2$. Squeeze...
11h
comment a simple question about inequality
Both $\sqrt{n+1}$ and $\sqrt{n+2}$ exceed $\sqrt n$. (Your inequality is in the wrong direction.)
12h
comment Find a Banach space E, Nearest point projection in uniformly convex Banach spaces
This post has one example. See the comments there, also.
14h
comment How to express the equations as the Square root Like?
This may help.
15h
comment prov of orthogonality of even functions and odd function
Is the product odd?
1d
comment Check proof about range of bounded linear operator.
Looks perfect now!
1d
comment Check proof about range of bounded linear operator.
Except, you should say "this clearly isn't the case, since $x$ is a sequence increasing to $\infty$".
1d
comment Check proof about range of bounded linear operator.
Did you mean to say $x\in\ell_\infty$? Your proof looks fine, other than that typo.
1d
comment If $\lim_{x\to\infty}\int_0^x a(t)\,dt=L$, is it true that $\lim_{x\to\infty}a(x)=0?$
The integral could "act" like a convergent alternating series. E.g., $\int_0^\infty\cos x^2\,dx$ converges.
1d
comment Kuratowski Theorem on uniform convergence
I think you're looking for Theorems 3-5, here (where "continuous convergence" is your condition). The terms in the link are defined on pages 197 and 201 of the same book.
1d
comment If [x,z] = 0 $\implies$ [x,y] = 0, then y = $\alpha$z. True for infinite dimensional vector space?
Perhaps this is a better link. You need to know the codimension of a non-zero linear functional on a vector space is $1$. For that, see this.
1d
comment If [x,z] = 0 $\implies$ [x,y] = 0, then y = $\alpha$z. True for infinite dimensional vector space?
It's true. See this.
1d
comment Constructing an example of a infinite set of triangles on the rational line whose union does not contain the interior of a rectangle.
@SomeStrangeUser For the Cantor set, I think any collection of triangles would be a counterexample. Any open line segment on the $x$-axis that intersects the Cantor set would contain one of the "deleted intervals" from the construction of the Cantor set. It would thus contain two points $a$ and $b$ from the Cantor set that are endpoints of one of the deleted intervals. The triangles corresponding to $a$ and $b$ would omit a portion of any, supposedly contained, rectangle.
2d
comment Function whose third derivative is itself.
See this.
2d
answered Constructing an example of a infinite set of triangles on the rational line whose union does not contain the interior of a rectangle.
2d
comment Constructing an example of a infinite set of triangles on the rational line whose union does not contain the interior of a rectangle.
Enumerate the rationals $\{r_i\}_{i=1}^\infty$. Let $T_1$ be arbitrary with bottom vertex $(0,r_1)$. Let $T_2$ be a triangle disjoint from $T_1$ with bottom vertex $(0,r_2)$. If $n$ pairwise disjoint triangles, $T_1,\ldots, T_n$ have been constructed with bottom vertices $r_1,\ldots, r_n$, it seems clear that a desired $T_{n+1}$ can be constructed. (Of course, I would not want to do this explicitly.)
2d
comment to evaluate using L'hopital's rule
@godonichia You could use the general rule mentioned by others; but I doubt that's the intended solution. Just use the elementary arguments presented here (in fact, this is easier).
2d
comment to evaluate using L'hopital's rule
Why does most every one appeal to the usual sledgehammer here, instead of taking the more edifying approach in this answer?
2d
comment Constructing an example of a infinite set of triangles on the rational line whose union does not contain the interior of a rectangle.
I may be off, but couldn't you inductively define the triangles so that the resulting collection is pairwise disjoint?
1 2 3 4 5