Joy of Mathematics

Rational limit

September 3, 2020 / Leave a comment

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,

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

Ordering values

August 29, 2020 / Leave a comment

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,

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

Double inequality

August 24, 2020 / Leave a comment

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

August 23, 2020 / Leave a comment

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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