Rational limit

Let \mathbb{N} \ni n \geq 2. Evaluate the limit

    \[\ell = \lim_{x \rightarrow 1} \frac{\left ( 1-\sqrt{x} \right )\left ( 1-\sqrt[3]{x} \right )\left ( 1-\sqrt[4]{x} \right ) \cdots \left ( 1-\sqrt[n]{x} \right )}{\left ( 1-x \right )^{n-1}}\]

Solution

Let us consider the function f_n(x) = \sqrt[n]{x}. Then,

    \[\frac{1}{n} = f'_n(1) = \lim_{x\rightarrow 1} \frac{f_n(x) - f_n(1)}{x-1} = \lim_{x\rightarrow 1} \frac{1-\sqrt[n]{x}}{1-x}\]

Hence,

    \[\ell = f'_2(1) \cdot f'_3(1) \cdots f'_n(1) = \frac{1}{2} \cdot \frac{1}{3} \cdots \frac{1}{n} = \frac{1}{n!}\]

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Ordering values

The following figure depicts the graph of the derivative of f.

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Consider the function g(x) = 2f(x) -x^2. Order the numbers g(-2), g(2), g(4).

Solution

We are working on the following figure.

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The area included by f' , the axis x'x and the lines x=2 , x=4 is less than \mathrm{E}_1. Hence,

    \begin{align*} \int_{2}^{4} \left |f'(x) \right | \, \mathrm{d}x < \mathrm{E}_1 & \Leftrightarrow \int_{2}^{4} f'(x) \, \mathrm{d} x < \frac{\left ( 4+2 \right ) \cdot 2}{2} \\ &\Leftrightarrow f(4) - f(2) < 6 \\ &\Leftrightarrow f(2) < f(4) - 6 \end{align*}

On the other hand \int \limits_{-2}^{4} f'(x) \, \mathrm{d}x >6. Hence,

    \begin{align*} \int_{-2}^{4} f'(x) \, \mathrm{d}x > 6 & \Leftrightarrow f(4) - f(-2) > 6 \\ &\Leftrightarrow f(-2) < f(4) -6 \end{align*}

Hence,

    \begin{align*} f(-2)<f(4)-6<f(2)&\Leftrightarrow 2f(-2)<2f(4)-12<2f(2) \\ &\Leftrightarrow 2f(-2)-4<2f(4)-16<2f(2)-4 \\ &\Leftrightarrow g(-2)<g(4)<g(2) \end{align*}

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Double inequality

Prove the following double inequality, where the sum and product are cyclic over the angles A, B, C of a triangle

    \[\sum \sin^2 A \leq 2 + 16 \prod \sin^2 \frac{A}{2} \leq \frac{9}{4}\]

Inscribed quadrilateral

Given a cyclic quadrilateral ABCD inscribed in a semicircle of diameter CD as shown at the figure with CD = x and sides of lengths a, b, c and x

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show that:

    \[x^3 - \left ( a^2 + b^2 + c^2 \right ) x -2abc =0\]

Solution

We recall Ptolemy’s theorem.

If a quadrilateral is inscribable in a circle then the product of the lengths of its diagonals is equal to the sum of the products of the lengths of the pairs of opposite sides.

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Applying Pythagoras’  Theorem to both ADC and DBC along with Ptolemy’s Theorem we get the result.

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Limit with harmonics

Let \mathcal{H}_n denote the n-th harmonic number. Evaluate the limit

    \[\ell = \lim_{n \rightarrow +\infty} \left ( \mathcal{H}_n - \frac{1}{n} \sum_{k=1}^{n} \mathcal{H}_k \right )\]

Solution

It follows from Cesaro and the definition of the Euler Mascheroni constant that the RHS is equal to

    \[\left ( \mathcal{H}_n -\ln n \right ) - \frac{1}{n} \sum_{k=1}^{n} \left ( \mathcal{H}_k -\ln k \right )+ \ln n - \frac{1}{n} \ln n!\]

The first two terms tend to \gamma - \gamma =0. All that is left to evaluate the limit of \ln n - \frac{\ln n!}{n} which equals 1 from here.

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