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5

Hint: Write the limit as: $\displaystyle\lim_{x \to 0^+}\dfrac{\tan^{-1}(e^x+\tan^{-1}x)-\tan^{-1}(e^{\sin x}+\tan^{-1}\sin x)}{x - \sin x} \cdot \dfrac{x-\sin x}{x^3}$ $= \displaystyle\lim_{x \to 0^+}\dfrac{\tan^{-1}(e^x+\tan^{-1}x)-\tan^{-1}(e^{\sin x}+\tan^{-1}\sin x)}{x - \sin x} \cdot \lim_{x \to 0^+}\dfrac{x-\sin x}{x^3}$ Use the mean value theorem ...


4

HINT Use the Fundamental Theorem of Calculus Part 1 which states $$\frac{d}{dx} \int^{h(x)}_{g(x)} f(t) \, dt = f(h(x))\cdot h'(x) - f(g(x))\cdot g'(x)$$ HINT 2


4

Although it's pretty obvious that $x(x+1)=x^2+x$ goes to infinity for $x\to\infty$, you can use partial fraction decomposition if it is not clear: $$\lim_{x\to\infty}{1\over x(x+1)}=\lim_{x\to\infty}\left({1\over x}-{1\over x+1}\right)=0$$


3

The first equality below uses the Fundamental Theorem of Calculus. The chain rule is used accordingly. \begin{align} \frac{dG}{dx} = \frac{1}{(x^2)^2+4} \cdot \frac{d}{dx}x^2=\frac{2x}{x^4+4} \end{align}


3

Continuing from O.L.'s answer, the following is an evaluation of $$\int_{0}^{\infty} \frac{\sin 2x}{x} \text{Ci}(x) \ dx .$$ First notice that by making the substitution $ \displaystyle u = \frac{t}{x}$, $$ \text{Ci}(x) = - \int_{x}^{\infty} \frac{\cos t}{t} \ dt = - \int_{1}^{\infty} \frac{\cos xu}{u} \ du.$$ Therefore, $$ \int_{0}^{\infty} \frac{\sin ...


3

Yes correct. Notice that you can also (and it's more simple) use the asymptotic equivalence of the function at $0$ and at $+\infty$. In fact, we have $$\frac{\arctan^2x}{x^2}\sim_\infty\frac{\pi^2}{4x^2}\in L^1([1,+\infty))$$ and $$\frac{\arctan^2x}{x^2}\sim_01\in L^1((0,1])$$ so the given integral is convergent.


3

We can use residue calculus here to find the integral value: $$ \int_{-\infty}^{\infty}e^{-i \xi x}f(x)dx = \int_\gamma e^{-i \xi z} \cdot \frac{z}{(z^2 + 4)^2}dz $$ which have double poles at $z = \pm 2i \Rightarrow z_1 = (z-2i)^{-2}, z_2 = (z + 2i)^{-2}$. We need to find the residues for $$ e^{-i \xi z}f(z) = (z - 2i)^{-2}ze^{-i \xi z}(z + 2i)^{-2} = ...


3

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} ...


3

The reasoning $$ \lim_{h \to 0}\cos(a + h) = \lim_{h \to 0}(\cos a\cos h -\sin a \sin h )= \cos a\cos 0-\sin a\sin 0=\cos a $$ reduces continuity of the cosine function to the continuity of the sine and the cosine at $0$, because only if you already know that the functions are continuous at zero you can say $$ \lim_{h\to 0}\cos h=\cos 0 ...


2

As it's a little complicated with finding the indefinite integral of $$\int\int(x^2+y^2)^{3/2}dydx$$ I would suggest using polar, as reversing the order of integration isn't really going to make a huge difference. Also the graph is a huge indicator of that.


2

Since you have $x^2 + y^2$, I suggest using polar coordinates. Since $$r^2 = x^2 + y^2$$ $$1 < r^2 < 4$$ So, $$1 < r < 2$$ To find $\theta$ $$\pi/6 < \theta < \pi/3$$ Don't forget to multiply $$rdrd\theta\ \text{in your double integral}$$


2

I use the definition that the derivative of a differentiable function $f$ at $x \in \mathbb{R} $ is: $$ f'(x) = \lim_{h \rightarrow 0 } \frac{f(x+h) - f(x)}{h} $$ Specifically this implies: $$ \forall \epsilon > 0 \ \ \ \ \ \exists n \in \mathbb{N} \ \ \ \left| \frac{f(x+1/n) - f(x)}{(1/n)} - f'(x) \right| < \epsilon $$ Define $ \log(b) $ as $ ...


2

Use the substitution $x = 3\sin \theta$. Then $dx = 3\cos \theta$. Your integral will reduce to $$\int \frac {3\cos \theta}{3\cos \theta}\,d\theta = \int 1\cdot \,d\theta=\int d\theta = \theta+C$$


2

Performing trig. substitution of $x=3 \sin \theta$ (so $dx=3\cos \theta \, d\theta$) gives $$\int\frac{1}{\sqrt{9-x^2}} \, dx=\int\frac{1}{\sqrt{9-9 \sin^2 \theta}} \, (3 \cos \theta \, d\theta)=\int\frac{1}{|3 \cos \theta|} \, (3 \cos \theta \, d\theta) = \int 1 \, d\theta$$ which is equal to $\theta + C$, or $\displaystyle\sin^{-1} \left( \frac{x}{3} ...


2

Well, we assume that you're considering Riemann integral, which is equivalent to Darboux integral. Suppose that $\int_a^bf=0$ and $f\ge0$, we'd show that $\exists x\in[a,b],f(x)=0$. Suppose $I$ is a closed interval, and $\int_I f=0$. By definition of Darboux integral, given $\epsilon>0$, there's a partition $x_0<x_1<\dotsb<x_n$ such that ...


2

I assume you have to use the $\epsilon$-$N$ definition of the limit instead of other methods, correct? If so, you shouldn't be trying to find an $\epsilon$. For a given $\epsilon > 0$, you need to find an $N$ such that $\left|\dfrac{(n+8)(n+1)}{n(n-10)}-1\right| < \epsilon$ for all $n > N$. So, you want $\left|\dfrac{19n+8}{n^2-10n}\right| ...


2

The function ${\displaystyle {x \over (x^2 + 4)^2}}$ is ${\displaystyle -{1 \over 2}}$ times the derivative of ${\displaystyle {1 \over x^2 + 4}}$, so its Fourier transform will be ${\displaystyle -{i \xi \over 2}}$ times the fourier transform of ${\displaystyle {1 \over x^2 + 4}}$. This one is standard and is given by ${\displaystyle {\pi \over 2} ...


2

One can rather easily compute the integral by residues. For example, for $\xi\geq0$ we can write \begin{align*} \mathfrak{F}(\xi)=-2\pi i\cdot \operatorname{res}_{z=-2i}\frac{ze^{-i\xi z}}{(z-2i)^2(z+2i)^2} =-2\pi i \cdot\left(\frac{ze^{-i\xi z}}{(z-2i)^2}\right)'_{z=-2i}=-\frac{\pi i}{4}\xi\, e^{-2\xi}. \end{align*} But since the answer should be an odd ...


2

$(\sin x)^x \to x \log \sin x = \frac{\log \sin x}{\frac{1}{x}}$


2

Hint: Use the Chain Rule and the Fundamental Theorem of Calculus.


2

Write it as an equation by appending $=0$ Multiply by $x^{3/4}$, move one term to the other side, and raise both sides to a convenient power.


2

You know that differentiability implies continuity, so let's just do (b) and (a) will be automatic. The tangent line to the function $y=\sin x$ at the point $x=\pi$ is given by $$y-\sin(\pi)={d\over dx}(\sin x\big|_{x=\pi}\cdot(x-\pi)$$ i.e. $$y-0=\cos(\pi)(x-\pi)\iff y=\pi-x$$ so that $m=-1$ and $b=\pi$.


2

I assume you mean you want this function to be continuous and differentiable at $x=\pi$. In order for the function to be continuous, you need $\sin(\pi)=m\pi+b$ so that $\displaystyle{\lim_{x\rightarrow\pi^-}y(x)=\lim_{x\rightarrow\pi^+}y(x)}$. For differentiability, you need $\displaystyle{\lim_{x\rightarrow\pi^-}y'(x)=\lim_{x\rightarrow\pi^+}y'(x)}$, so ...


1

Hint: What does it mean to be continuous? It means that $$f(\pi)=\lim_{x \rightarrow \pi}f(x)$$. Take limits from the left and right of $\pi$. For a function to have a derivative at a point requires it to be also continuous at that point. Thus, you can use the information you found from continuity. Use similar reasoning as above for the limit of the ...


1

If we have$$\frac{1}2 x^{-1/2}- \frac14x^{-3/4}=0$$ then we multiply both sides by $x^{3/4}$ to get $$\frac 12 x^{1/4}-\frac 14=0.$$ Solving for $x$ gives $x=\frac{1}{16}$.


1

We can proceed as follows $$\begin{aligned}L\, &= \lim_{x \to 0^{+}}\frac{\tan^{-1}(e^{x} + \tan^{-1}x) - \tan^{-1}(e^{\sin x} + \tan^{-1}(\sin x))}{x^{3}}\\ &= \lim_{x \to 0^{+}}\dfrac{\tan^{-1}\left(\dfrac{e^{x} + \tan^{-1}x - e^{\sin x} - \tan^{-1}(\sin x)}{1 + (e^{x} + \tan^{-1}x)(e^{\sin x} + \tan^{-1}(\sin x))}\right)}{x^{3}}\\ &= \lim_{t ...


1

Incomplete answer (maybe it can be used): If $\left|z\right|<1$ then $\frac{1}{1-z}=1+z+z^{2}+\cdots$ Applying this for $z=x^{4},x^{3},x^{2}$ you find: $$\frac{1}{\left(1-x^{4}\right)\left(1-x^{3}\right)\left(1-x^{2}\right)}=\left(1+x^{4}+x^{8}+\cdots\right)\left(1+x^{3}+x^{6}+\cdots\right)\left(1+x^{2}+x^{4}+\cdots\right)$$ Then $a_{n}$ must equalize ...


1

Hints : First, prove that $$\frac{1}{(1-x^2)(1-x^3)(1-x^4)} = \frac{7}{32(x+1)}-\frac{59}{288(x-1)}+\frac{1}{8(x-1)^2}+\frac{1}{16(x+1)^2}-\frac{1}{24(x-1)^3}+\frac{x+2}{9(x^2+x+1)}+\frac{1-x}{8(x^2+1)}.$$ Then, use that $$ \frac{1-x}{x^2+1} = -\frac{1+i}{2(x-i)}+\frac{-1+i}{2(x+i)}.$$ and $$ \frac{x+2}{x^2+x+1} = ...


1

The $n$-th term is obtained from $(1+x^2+x^4+\cdots)(1+x^3+x^6+\cdots)(1+x^4+x^8+\cdots)$ which is the number of triples $(i,jk)$ that are solutions to $2i+3j+4k=n$. Hence, $$a_n=\sum_{2i+3j+4k=n}1=\sum_{j=0}^{\lfloor n/3\rfloor }\sum_{2i+4k=n-3j}1$$ Now we want to obtain $$b_m=\sum_{2i+4k=m}1$$ It is clear $b_{2m+1}=0$, so we look at ...


1

As a similar and useful hint, one can use the definition of Infinitesimal Functions here. In fact, while $x\to 0$ we have $\sin x\sim x$ . Now apply the method you know. I mean put $y=x^x$ and then...



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