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2

Note that as $x\to +\infty$, $$\cot^{-1}(x)=\arctan(1/x)=\frac{1}{x}-\frac{1}{3x^3}+o(1/x^3).$$ Hence, from your work, $$\frac{-x^3}{2}\bigg(\frac{-x}{1+x^2} + \cot^{-1}(x)\bigg)= \frac{-x^3}{2}\bigg(-\frac{1}{x}\frac{1}{1+\frac{1}{x^2}} + \frac{1}{x}-\frac{1}{3x^3}+o(1/x^3)\bigg)\\ =\frac{-x^3}{2}\bigg(-\frac{1}{x}+\frac{1}{x^3} + \frac{1}{x}-\frac{1}{3x^3}...


2

All that you have proved was that if your limit exists, then it cannot be greater than $0$. You can prove that it is $0$ using the fact that$$\left\lvert(x^2-y^2)\sin\left(\frac1{x^2+y^2}\right)\right\rvert\leqslant\lvert x\rvert^2+\lvert y\rvert^2.$$


2

$$ \lim_{x\to 1}\frac{P(x)}{(x-1)^2}\hbox{ exists}\implies\lim_{x\to 1}P(x) = 0\implies P(1) = 0\implies P(x) = (x - 1)Q(x)\hbox{ with $Q(x)$...} $$ Can you say why all the $\implies$ are true? Can you continue?


2

Every point on the limiting curve has a distance 1 to the $x$-axis along the tangent line at that point. Intuitively, this should be plausible. More formally, let $x=f(y)$ be the equation for the curve. The equation for the tangent line at $(f(y_0),y_0)$ is $$x-f(y_0)=f'(y_0)(y-y_0).$$ This meets the $x$-axis at $(y_0f'(y_0)+f(y_0),0)$ The condition then ...


1

By the triangle inequality, $$d(y_{m_k},z_{m_k}) \leq d(y_{m_k},y) + d(y,z) +d(z,z_{m_k})\tag{1}$$ and also $$d(y,z) \leq d(y, y_{m_k}) + d(y_{m_k},z_{m_k}) +d(z_{m_k},z)\tag{2}$$ Combining $(1)$ and $(2)$ gives $$|d(y,z)-d(y_{m_k},z_{m_k})|\leq d(y, y_{m_k}) +d(z_{m_k},z)$$ Now, the right-hand side converges to zero, by assumptions.


1

We have that $$\text{arccot}{(x)}=\arctan{\left(\frac1x\right)}=\frac1x-\frac1{3x^3}+o\left(\frac1{x^3}\right)$$ as $x\to\infty$. Also note that as $y\to0$ we have $$\ln{(1+y)}=y+o(y)$$ Hence our limit is simply $$\begin{align} \lim_{x\to\infty}x^2\ln{(x\text{arccot}{(x)})} &=\lim_{x\to\infty}x^2\ln{(x(1/x-1/(3x^3)+o(1/x^3)))}\\ &=\lim_{x\to\infty}x^...


1

HINT Letting $x=a^3$ the expression becomes $$\dfrac{e^x}{(e^x-1)/x}$$


1

Clearly, when $a>0$, $0\leq 1-x^a\leq 1$ for $x\in[0,1]$ and $\max_{x\in[0,1]}(1-x^a)=1-0=1$. So $\left(\displaystyle \int_0^1(1-x^a)^n\,dx\right)^{\frac1n}\leq 1.$ Since $1-x^a$ is continuous, for every $0<\epsilon<1$ we can find $\delta>0$ such that $1-\epsilon\leq 1-x^a$ for $x\in[0,\delta]$, hence $$\left(\int_0^1(1-x^a)^n\,dx\right)^{\...


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