# Tag Info

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


Only top voted, non community-wiki answers of a minimum length are eligible