11h
comment Exists $x_{n}$ such that $f'(x_{n})\rightarrow f'(x)$?
@user159517 I'd assume the problem called for $f$ differentiable only; as otherwise it's trivial, as you point out.
12h
comment Exists $x_{n}$ such that $f'(x_{n})\rightarrow f'(x)$?
Look at secant lines that converge to the tangent line. Use the Mean Value Theorem.
13h
comment What does it mean to "admit" something in vector calculus?
It means, "if there exists a vector potential $A$..."
14h
comment Suppose that $ T$ is a linear operator from a normed space $ X $ into normed space $ Y $.Prove that the following are equivalent
The answer here may help.
1d
comment No pairs when drawing cards from deck
In your first display, you have ordered selections upstairs and unordered downstairs.
1d
comment Solve by Using the Squeeze Theorem
I'm not sure what you mean. (I didn't multiply an inequality by a negative. I just used the fact mentioned at the start of my previous comment.)
1d
comment Solve by Using the Squeeze Theorem
No. $|x|\le a$ if and only if $-a\le x\le a$. Let $A$ be your limit expression. Then $|A|\le \sqrt{x^3+x^2}$; so, $-\sqrt{x^3+x^2}\le A\le \sqrt{x^3+x^2}$.
1d
comment Solve by Using the Squeeze Theorem
Use $-\sqrt{x^3+x^2}$ and $\sqrt{x^3+x^2}$.
1d
comment Solve by Using the Squeeze Theorem
Hint: $|\sin\theta|\le1$.
2d
comment Give an example of an operator that ${\cal D}(A)=X=l_1$, $Y=l_2$?
$x\mapsto 0$. There are more interesting examples, such as the formal identity operator.
2d
comment If $f $ is a continuous function then $\exists a$ such that..
Hint: if $m$ is the minimum of $f$ over $[x,y]$ and $M$ is the maximum, then $m\le{1\over y-x}\int_x^yf(s)\,ds\le M$.
2d
comment Let $f: \Bbb R \to \Bbb R$ be such that $f^{-1} (a, \infty)$ and $f^{-1} (- \infty, b)$ are open for any $a,b \in \Bbb R$.
Use what you've done to show that for $O$ open, $f^{-1}(O)$ can be written as a union of open sets (and is, thus, open).
2d
comment Let $f: \Bbb R \to \Bbb R$ be such that $f^{-1} (a, \infty)$ and $f^{-1} (- \infty, b)$ are open for any $a,b \in \Bbb R$.
An open subset of $\Bbb R$ need not be an interval.
2d
comment Theorem 4.20(c) in Baby Rudin: Is every continuous function whose domain is an unbounded subset of $\mathbb{R}$ uniformly continuous?
Whatever Theorem 4.20(c) is, the answer is "no".
2d
comment Why if $T$ is not a bounded operator then exists $ (x_n) $ that converges to $ 0_{X} $ for which $ \| T(x_n) \| \geq n^2 $ for all $ n $?
You can't. But, there is a sequence $x_n$ of norm-one elements with $\Vert T x_n\Vert>n^3$. Consider the sequence $(x_n/n)$.
2d
revised Prove $\lim_{n\to \infty}b_n=\infty$.
deleted 6 characters in body
Feb
23
comment Find integral $\int\frac{1}{cosx}dx$
See this.
Feb
21
comment Use continuity to evaluate the $\lim_{x\to\pi}(7\sin(x + \sin x))$
No. You just note (or show) that your function is continuous at $x=\pi$. The definition of continuity at a point tells you the limit is just the function evaluated at $\pi$. See Hector's nice answer.
Feb
21
comment Use continuity to evaluate the $\lim_{x\to\pi}(7\sin(x + \sin x))$
It basically means just "plug pi into x".
Feb
20
comment passing to a liminf from weak convergence.
See this.
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