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4

Hint: replace $1+xy$ by $1$ in the numerator (why?) and reduce to $$ \lim_{(x,y) \to (0,0)} \left( \frac{x^3 y^3}{x^3+y^3} \right)^{1/3}. $$ Put $\alpha =x^3$, $\beta=y^3$ and consider the constraint $\alpha + \beta = \alpha^3$ with $\alpha \to 0$. Then $$ \frac{\alpha \beta}{\alpha+\beta} = \frac{\alpha^4-\alpha^2}{\alpha^3} $$ which becomes unbounded as ...


4

Let we set $A=x^2,B=y^2,C=z^2$ and: $$ S_n(A,B,C) = \frac{A^n}{(A-B)(A-C)}+\frac{B^n}{(B-A)(B-C)}+\frac{C^n}{(C-A)(C-B)} .\tag{1}$$ By partial fraction decomposition, it is straightforward to check that $S_0=S_1=0$ and $S_2=1$. For every $n>2$ induction gives: $$ S_n(A,B,C) = h_{n-2}(A,B,C)\tag{2} $$ where $h_k$ is a complete homogeneous symmetric ...


3

You have to distinguish between the image of your map $r$ and its parametrization. Since your definition of surface does not include the requirement of injectivity of $r$ the image set may have self intersections, and whereever that happens, the image may have several tangent planes, in particular if the self intersectios are transversal. Apart from that, ...


3

Divide top and bottom by $x^2$ to get rid of the sine, as $\sin x^2 / x^2 \to 1$, and take $y$ out of the new numerator to end up with $$\lim_{(x,y)\to(0,0)}y\frac{x/y + y^2 / x^2}{1 + y^2 / x^2}, $$ which "could" not exist or be non-zero only if $\lvert x/y \rvert \to \infty$, but then again the limit would be the same as $$ \lim_{(x,y)\to(0,0)}y ...


3

From calculus of one variable it is known that $$\lim_{t\to 0}\frac{\sin t}{t}=1$$ Then, if $y\neq 0$ let $t=xy$, $x=t/y$, we have \begin{align*} \lim_{x\to 0}\frac{\sin (xy)}{x}&=\lim_{x\to 0}\left[\frac{y}{1}\frac{\sin (xy)}{xy}\right]\\ &=y\lim_{t\to 0}\frac{\sin t}{t}\\ &=y(1)\\ &=y \end{align*} Then, in order to get $f$ be continuous ...


3

The statement is a bit vague but I think this is probably what the author means. The graph of a function of two variables... is a level set of a function of three variables. That is, if $f(x,y)$ is a function of two variables, then the graph $z=f(x,y)$ can be written as $$z-f(x,y)=0\ .$$ If you now define a function of three variables $$g(x,y,z)=z-f(x,y)\ ...


3

A circle, i.e, an equation of the form $(x-a)^2+(y-b)^2=r^2$ is a relation on $\Bbb R$, that is, a subset of $\Bbb R^2$, which relates an element x with an element y, with some criteria, in this case, the circle equation. To write it explicitly, the relation would be $C=\left\{(x,y)\in\Bbb R^2: (x-a)^2+(y-b)^2=r^2\right\}$.


3

It makes sense to add two points from a vector space. It doesn't really make sense to add two points from an affine space. That is, you need to impose a system of coordinates before it makes sense to add them. (If you remove some of the structure from a vector space, you get an affine space.) Originally, the author was probably talking about $\mathbb{R}^n$ ...


2

Why do we need the linearity of $L$ wrt $h$? What do we lose if $L$ is not linear? The derivation at a point shall be the best linear approximation of the function at this point. More precisely $L$ is the best linear approximation of the difference function $g(h):=f(x+h)-f(x)$ for a fixed point $x$. Why $x$ is the argument of $L$, isn't it fixed? Shouldn't ...


2

We are given $\nabla f=C\vec r$, where $\vec r$ is the position vector and $C$ is the constant of proportionality. Then, by integration, we find that $f$ is given by $$f(x,y,z)=\frac12 C(x^2+y^2+z^2)+C'$$ where $C'$ is an integration constant. Evaluating $f$ at $(0,0,a)$ and $(0,0,-a)$ reveals that $$f(0,0,a)=\frac12Ca^2+C'$$ and $$\begin{align} ...


2

To show that $F$ is not conservative on $\mathbb{R}^2 \setminus \{ 0 \}$, compute the curve integral of $F$ along the unit circle. It will turn out to be non-zero, which shows $F$ can't be conservative. For b), the domain $\Omega$ is simply connected, so the fact that $\operatorname{rot}(F) = 0$ on $\Omega$ is enough to guarantee that $F$ is convervative ...


2

The induced metric is just the Euclidean metric on $\mathbb R^{n+1}$, i.e. $g_{ij} = \delta_{ij}$, but restricted to act on vectors tangent to $S^n$. You need to choose a system of $n$ coordinates parametrizing $S^n$ alone (or an $n$-frame tangent to $S^n$) if you want to get an $n \times n$ matrix of components for $g$.


2

In a multivariate environment, you can't guarantee that the answer you get using L'Hopital will make your function continuous. However, you can guarantee that if your function can be made continuous, L'Hopital's rule will give you the right answer. In your case, you want $f(x,y)$ to be continuous on $\mathbf{R}^2$. That means, in particular, if you hold $y$ ...


2

You can use the fundamental limit for $\sin$ instead: $$\frac{\sin(xy)}{x} = y\frac{\sin(xy)}{xy} \stackrel{x \to 0}{\longrightarrow} y \cdot 1 = y.$$The point is that $y$ is fixed here.


2

I would use the squeeze theorem. Since in a neighbouhood of $0$ you have $$\sin 4t^2 \ge t^2$$ you can substitute $x=2t$ and get $$0 \le \frac{x^3+y^3}{\sin x^2+y^2} = \frac {8t^3+y^3}{\sin 4t^2+y^2} \le \frac {8t^3+y^3}{t^2+y^2} \to 0$$ so the limit is $0$.


2

The tangent plane and normal are unique as defined, and also invariant in isometric mappings.


2

HINT: Note that we have $$\frac{\partial g}{\partial \nu}=\frac{2}{\epsilon}$$ on $\Gamma_{\epsilon}$. Therefore, $$\lim_{\epsilon\to 0}\int_{\Gamma_{\epsilon}}\left(g\frac{\partial\phi}{\partial\nu}-\phi\frac{\partial g}{\partial\nu}\right)\,ds=-4\pi\phi(0)$$


2

Since \begin{eqnarray} \frac{\partial F}{\partial x}(0,0)&=&\frac{\partial f}{\partial x}(0,0)+2\frac{\partial f}{\partial y}(0,0)=\nabla f(0,0)\cdot(1,2)=(4,-3)\cdot(1,2)=-2\\ \frac{\partial F}{\partial y}(0,0)&=&3\frac{\partial f}{\partial x}(0,0)-\frac{\partial f}{\partial y}(0,0)=\nabla f(0,0)\cdot(3,-1)=(4,-3)\cdot(3,-1)=15, ...


2

Given a function $f:\mathbb R^n \to \mathbb R^m$ and $h\in\mathbb R^n$. The directional derivative at $0$ in direction $h$ is defined by (if the limit exists) $$ f'(0; h) := \lim_{t\to 0} \frac{f(th) - f(0)}{t}. $$ The function, $h\mapsto f'(0; h)$ is called the Gâteaux differential or derivative. Notice that $f'(0;h)$ satisfies the linear approximation ...


1

It seems to me that if you write $f(x,y,z)=y^2{x}-y^2+z^2=0$ as $y=\sqrt{ \frac{z^2}{1-x}}$ then you have $y(x,z)$. Then: $$ \frac{\delta{f}(x,y,z)}{\delta{x}}=y^2 $$ $$ \frac{\delta{f}(x,y,z)}{\delta{y}}=2x{y}-2y $$ $$ \frac{\delta{y}(x,z)}{\delta{x}}=\frac{z}{2}{(\frac{1}{1-x})}^{\frac{3}{2}} $$ then: $$ ...


1

If this limit really exists, then so should the limit of each variable, one at a time. This is both a necessary and sufficient condition for the existence of the overall $\mathbb{R}^2$ limit. Look at $\lim \limits_{y \to 0} \lim \limits_{x \to 0} \frac{x^3+\sin(x^2+y^2)}{y^4+\sin(x^2+y^2)} = \lim \limits_{y \to 0} \frac{\sin(y^2)}{y^4 + \sin(y^2)}$. Now ...


1

$f(x,y)$ isn't discontinuous at $(0,0)$. How do I know? First of all, can you get a picture of this function in your head? I can. Since $f: \Bbb R^{2} \to \Bbb R$, you should think of this function as first taking the $XY$-plane in $3$D space (let it be the horizontal axes), and warp that plane. In this case, we are warping the plane like a taco. If ...


1

$\forall h\neq 0,\quad \left | \frac{f(0,0+h)-f(0,0)}{h} \right |<\left | \frac{h^{2}}{h} \right |=\vert h\vert $ so if we let $h\to 0$, we see that $f_y(0,0)=0$. Likewise, $\forall h\neq 0,\quad \left | \frac{f(0+h,0)-f(0,0)}{h} \right |=\left | \frac{0}{h} \right |=0$ so $f_x(0,0)=0$. Now, $\vert f(x,y)-f(0,0)\vert <y^{2}$ which is $<\epsilon ...


1

No, no and no: they are very different things. The derivative (also called differential) is the best linear approximation at a point. The directional derivative is a one-dimensional object that describes the "infinitesimal" variation of a function at a point only along a prescribed direction. I will not write down the definitions here. So to speak, the ...


1

After calculating $P_x,Q_y,R_z$, deduce that $\nabla \cdot F$ (the sum of these) is zero, which means (by the divergence theorem) that the total flux (which is an integral of $\nabla \cdot F$) is zero. There's nothing wrong with your reasoning. You could calculate the flux without the theorem with a parametrized surface integral, but this approach would be ...


1

Since you are dealing with a moment of inertia, the claim just follows from the Huygens-Steiner/parallel axis theorem: the absolute minimum is attained by the centroid of $\{a_1,\ldots,a_m\}$. On the other hand, your argument is perfectly fine, but you do not need to compute any Hessian matrix, since the moment of inertia is a convex function as a sum of ...


1

Your computation of $x_0$ is correct, except that you forgot to divide through $m$ in the end. ($x_0$ is the center of gravity of the $a_k$.) The argument that $\lim_{|x|\to\infty} f(x)=\infty$ is correct. Then you can, for example, argue that for a sufficiently large radius $r$, the function $f$ restricted to the closed ball of radius $r$ around $0$ has a ...


1

$f'(c,u)=\nabla f(c) . u=(4,-3). (\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})= 4\frac{\sqrt{2}}{2}-3\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{2}$ ( "." is inner product symbole.)


1

$f(x,y)= A_{produced}-A_{degrade}$ $A_{produced}(y)=\frac{n}{1+y}$ $A_{produced}(0)=3$ $3=\frac{n}{1+0}$ $n=3$ $A_{degrade}= x$ $f(x,y)= \frac{3}{1+y}-x$


1

If $R$ is large enough $\varphi$ and $\nabla \varphi$ are equal to $0$ on $\Gamma_2$, and so is the RHS in the last formula.



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