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12

Yes correct but simply you can do $$0\le \left|x\sin\left(\frac1x\right)\right|\le|x|$$ and use the squeeze thorem.

7

Choosing the substitution $x - \pi = t$, with $t \to 0$, we have $$\lim_{x \to \pi}\frac{\sin x}{x^2 - \pi^2} = \lim_{t \to 0}\frac{\sin(t + \pi)}{t(t + 2\pi)} = \lim_{t \to 0}-\frac{\sin t}{t(t + 2\pi)} = -\frac1{2\pi}$$

6

$$0 \le \frac{\sin^2 n}{2^n} \le \frac 1{2^n} \to 0$$ Use the squeeze theorem to conclude Note that $|x| \le 1 \implies x^2 \le 1$, so given that $-1\le \sin x \le 1$ you can conclude that $\sin^2 \le 1$ (also it is positive, but that because it is a square) In general you have $|\sin^k x| \le 1$

6

here is a slightly quicker way. take the inequality $-1 \le \sin \dfrac{1}{x} \le 1$ this gives $$-|x| \le x\sin \dfrac{1}{x} \le |x|$$ now use the squeeze theorem.

5

This is true for $a\in \mathbb Z$ too. $-(\pi-x)a$ is real, so $-i(\pi-x)a$ is purely imaginary. It is well known that $$e^{it}=\cos t + i\sin t \qquad\qquad\text{when }t\in\mathbb R$$ so its modulus is $\sqrt{\sin^2 t+\cos^2 t} = \sqrt 1 = 1$.

5

Do the substitution $\pi-x=t$, so your limit becomes $$\lim_{t\to0}\frac{\sin(\pi-t)}{-t(2\pi-t)}= \lim_{t\to0}\frac{\sin t}{-t(2\pi-t)}$$ which is elementary.

4

The image below might help to explain where this term comes from:

4

Your last integral $$4\int_{-1}^{1} (2x-3)\sqrt{1-x^2}\;dx$$ Can be split into the two integrals $$4\int_{-1}^{1} 2x\sqrt{1-x^2}\;dx + 4\int_{-1}^{1}(-3)\sqrt{1-x^2}\;dx$$ The second integral you can compute in terms of $\pi$, and the first integral can be computed via $u$-substitution.

3

Let $l_t$ be the horizontal line $y=t$. Now let $P$ denote the points on and inside the polygon. $$A(t)=\{(x,y)\in P: y\le t\}$$ and $$B(t)=\{(x,y)\in P: y\ge t \}.$$ Finally let $$d(t)=\text{Area }A(t)-\text{Area }B(t)$$. You can apply Intermediate Value Theorem to $d(t)$.

3

Simple way: since $(x^7-1)^2+1$ never vanishes, $f(x)$ is a continuous function on $\mathbb{R}$. Since $\lim_{x\to \pm \infty} f(x)=1000$, for some $M\in\mathbb{R}$ we have that $f(x)$ is between $999$ and $1001$ on the complement of $[-M,M]$. Since $I=[-M,M]$ is a closed interval and $f$ is a continuous function, $f$ is bounded on $I$. Since it is bounded ...

3

You did everything right except for one thing. If you'd like to find the position at $t \geq 1$, you start with $r(1)$ as you correctly did, and add $t-1$ times $r'(1)$. Why $t-1$ and not $t$? Because $t-1$ is exactly how much time passes since $t=1$. So the parametrization is \begin{align} r(t) &= r(1)+(t-1)r'(1)\\\\ ...

3

There does not exist an indefinite integral because of the intermediate value theorem for derivatives (Darboux's theorem). If you are given a concrete interval over which to integrate, the question is trivial, because the function is piecewise constant.

2

I assume that the line $L_1$ is the intersection of the planes given by the two equations that you write. The answers given so far assume that your given lines $L_1$ and $L_2$ intersect. However, the lines are parallel, so taking the product of their direction vectors just gives you the zero vector and is of no help. Instead, you could take a vector from ...

2

...Hint: $\dfrac{1}{\sqrt{t}+\sin t} \leq \dfrac{1}{\sqrt{t}}$

2

This example is given in Rudin's mathematical analysis text. Let $a = -1$ and set $$f(x) = \begin{cases}2x^2 - 1 & (-1 < x < 0)\\ \\\dfrac{x^2 - 1}{x^2 + 1} & (0 \le x < \infty)\end{cases}$$ Then $$f'(x) = \begin{cases}4x & (-1 < x < 0)\\ \\ \dfrac{4x}{(x^2 + 1)^2} & (0 \le x < \infty)\end{cases}$$ and $$f''(x) = ... 2 Hint Assume f(x) \not\equiv 0. Then there is some x' such that f(x') \ne 0. By MVT we get the existence of a \xi\in[0,x'] with$$|f(\xi)| \ge |f'(\xi)| = \left|\frac{f(x')}{x'}\right|$$Thus if |x'| < 1 we get that \xi < x' and f(\xi)\ne 0. Use this Ansatz to prove that f\equiv 0 in [0,1] and translate f to f(x+c) to conclude. 2 |\sin x|\sin(nx) is an odd function, so its integral on the interval [-\pi,\pi] is equal to 0. On the other hand |\sin x|\cos(nx) is even. Its integral on [-\pi,\pi] is twice the integral on [0,\pi]. The last integral should be from 0 to \pi. 2 We can factorize the expression as follows:$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1} = \lim_{x \to 1}\frac{(x - 1)(3x^3 - 5x^2 - 5x - 5)}{(x - 1)^2(x + 1)} = \lim_{x \to 1}\frac{3x^3 - 5x^2 - 5x - 5}{(x - 1)(x + 1)}$$Now consider the two limits from the sides:$$\begin{align}\lim_{x \to 1^-}\frac{3x^3 - 5x^2 - 5x - 5}{(x - 1)(x + 1)} = +\infty\\ ...

1

Your answer is the right idea, but you need a few more details. Let $c = \sup \{ t \in [a,b] | f(t) = f(a) \}$, then we have $c <b$ and $f(c) = f(a)$ by continuity. Furthermore, $f(x) > f(a)$ for all $x > c$. Now let $d = \inf \{ t \in [c,b] | f(t) = f(b) \}$. We have $c < d$ and $f(d) = f(b)$. Furthermore, $f(x) < f(b)$ for all $c \le x ... 1 This is a revision of my previous answer, which was not accurate enough. Let$f,g:\mathbb{R}^n\to\mathbb{R}^n$be two continuously differentiable functions, and let$\langle\cdot,\cdot\rangle$be some inner product. The Leibniz rule says that the directional derivative of$\langle f,g\rangle$at a point$x$in the direction$uis given byd\langle ... 1 The Cauchy condensation test may help here. For a non-increasing sequence (f(n))_{n\in\mathbb{N}} of non-negative real numbers we have (from Oresme's proof of the divergence of the harmonic series) : \begin{align} \tag{1}\sum_{k=1}^\infty f(k)&\le \underbrace{f(1)}_{f(1)}+\underbrace{(f(2)+f(2))}_{2\times ... 1 Here are the steps\int\frac{8}{x^2+49}dx=\frac{8}{49} \int\frac{1}{\frac{x^2}{49}+1}dx $$Let u=\frac{x}{7}, then du=\frac{1}{7}dx. So now$$ \frac{8}{7} \int\frac{1}{u^2+1}du= \frac{8}{7} \arctan u+C  = \frac{8}{7} \arctan\left(\frac{x}{7}\right)+C $$1 "So there exists a point c such that f(c)=0." That should be "There exists a point y\in(0,x) such that f(y)=c." Afterall c in the range value of f, and you want to find an y for c=0. 1 Note that since a_n \gt 0 and a_0 \lt 0, it must be the case that n \gt 0 to avoid an inconsistency. While it is clear that P(0) = a_0 \lt 0, the intuitive notion that P(x) attains positive values for all sufficiently large x deserves a little justification. What we have to work with is that the leading coefficient a_n \gt 0 is positive. ... 1 Here's some remarks: You said that "f(x) converges uniformly to some continuous f(x)" which is misspoke and we should write "the series converges uniformly..." You said "... Hence, we may take the derivative of the term...": I don't see the implication: how if f is continuous then you may take the derivative? Finally, notice that to prove \sum_n ... 1 It's not complete, because you only have computed$$ \lim_{x\to 0^+}x\sin\frac{1}{x} $$and you should also do the limit from the left; however, as \sin(-\alpha)=-\sin\alpha, the function is even, so the limit from the left is equal to the limit from the right, provided one of them exists. Your substitution, however, just exploits the fact that ... 1 Let a=a_1+ia_2, \>b=b_1+i b_2. Your map F_{a,b} can be written as$$F_{a,b}=\exp \>\circ\> T\ .$$Here$$T:\>{\mathbb C}\to{\mathbb C}, \qquad z\mapsto a\>{\rm Re}(z)+b\>{\rm Im}(z)$$can be considered as a real linear map$$T:\>{\mathbb R}^2\to{\mathbb R}^2,\qquad (x,y)\mapsto(a_1x+b_1y, \>a_2x+b_2 y)$$with matrix ... 1 The theorem tells you that$$\int_a^b f(x)\mathrm dx = F(b) - F(a)$$where F is any function that verifies F'(x) = f(x) for every x \in (a, b). You have already found such an F, namely$$F(x) = -\frac12\ln\cos2x$$If you really want to be sure that indeed F'(x) = f(x) you can just differentiate using the theorems of differentiation. But that is ... 1 Hint Let x=\dfrac{\pi}{2}+t,then$$\sin^n{x}\sin{(n+2)x}=\cos^{n}{t}\sin{[(n+2)(\frac{\pi}{2}+t)]}$$is odd function$$I=\int_{0}^{\pi}\sin^n{x}\sin{(n+2)x}dx=\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\cos^{n}{t}\sin{[(n+2)(\frac{\pi}{2}+t)]}dt=0$1$a_{N+2}\le ra_{N+1}$and since$a_{N+1}\le ra_N$,$ra_{N+1}\le r(ra_N)$. In other words,$a_{N+2}\le r^2a_N\$. Is this what you are asking about?

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