Common Tangents and Areas.

Level pending

Let n n be a positive integer, f ( x ) = x 1 2 n f(x) = x^{\frac{1}{2n}} and g ( x ) = − x 2 n g(x) = -x^{2n}

f ( x ) f(x) and g ( x ) g(x) have a common tangent at points A A and B B respectively.

Let A ( n ) A(n) be the area of the region R R bounded by the tangent line and f ( x ) f(x) and g ( x ) g(x)

Find lim ⁡ n → ∞ A ( n ) \lim_{n \rightarrow \infty} A(n) .


The answer is 0.5.

Rocco Dalto by Rocco Dalto , 67, USA

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1 solution

Rocco Dalto
Dec 29, 2019

Let f ( x ) = x 1 2 n f(x) = x^{\frac{1}{2n}} and g ( x ) = − x 2 n g(x) = -x^{2n}

From the graph above R ( 0 , 0 ) ∘ R y = x ( x 0 , y 0 ) = ( − y 0 , − x 0 ) R_{(0,0)} \circ R_{y = x} (x_{0},y_{0}) = (-y_{0},-x_{0})

Using A : ( x 0 , y 0 ) A: (x_{0},y_{0}) and B : ( − y 0 , − x 0 ) ⟹ B: (-y_{0},-x_{0}) \implies slope m A B = 1 ⟹ m_{AB} = 1 \implies

f ′ ( x 0 ) = 1 2 n ( 1 x 0 ) 2 n − 1 2 n ⟹ x 0 = ( 1 2 n ) 2 n 2 n − 1 ⟹ y 0 = ( 1 2 n ) 1 2 n − 1 ⟹ f'(x_{0}) = \dfrac{1}{2n}(\dfrac{1}{x_{0}})^{\frac{2n - 1}{2n}} \implies x_{0} = (\dfrac{1}{2n})^{\frac{2n}{2n - 1}} \implies y_{0} = (\dfrac{1}{2n})^{\frac{1}{2n - 1}} \implies

A : ( ( 1 2 n ) 2 n 2 n − 1 , ( 1 2 n ) 1 2 n − 1 ) A: ((\dfrac{1}{2n})^{\frac{2n}{2n - 1}}, (\dfrac{1}{2n})^{\frac{1}{2n - 1}}) and B : ( − ( 1 2 n ) 1 2 n − 1 , − ( 1 2 n ) 2 n 2 n − 1 ) ⟹ B: (-(\dfrac{1}{2n})^{\frac{1}{2n - 1}}, -(\dfrac{1}{2n})^{\frac{2n}{2n - 1}}) \implies

y = x + ( 2 n − 1 ) ( 1 2 n ) 2 n 2 n − 1 y = x + (2n - 1) (\dfrac{1}{2n})^{\frac{2n}{2n - 1}}

Let j ( n ) = ( 1 2 n ) 2 n 2 n − 1 ⟹ y = x + ( 2 n − 1 ) j ( n ) j(n) = (\dfrac{1}{2n})^{\frac{2n}{2n - 1}} \implies y = x + (2n - 1) j(n)

⟹ A 1 ( n ) = ∫ 0 j ( n ) ( x + ( 2 n − 1 ) j ( n ) − x 1 2 n ) d x = \implies A_{1}(n) = \displaystyle\int_{0}^{j(n)} (x + (2n - 1) j(n) - x^{\frac{1}{2n}}) \:\ dx =

x 2 2 + ( 2 n − 1 ) j ( n ) x − 2 n 2 n + 1 x 2 n + 1 2 n ∣ 0 j ( n ) = \dfrac{x^2}{2} + (2n - 1)j(n)x - \dfrac{2n}{2n + 1}x^{\frac{2n + 1}{2n}}|_{0}^{j(n)} =

j ( n ) 2 2 + ( 2 n − 1 ) j ( n ) 2 − 2 n 2 n + 1 j ( n ) 2 n + 1 2 n ∣ 0 j ( n ) = \dfrac{j(n)^{2}}{2} + (2n - 1)j(n)^2 - \dfrac{2n}{2n + 1}j(n)^{\frac{2n + 1}{2n}}|_{0}^{j(n)} =

( 4 n − 1 2 ) ( 1 2 n ) 4 n 2 n − 1 − ( 2 n 2 n + 1 ) ( 1 2 n ) 2 n + 1 2 n − 1 = (\dfrac{4n - 1}{2})(\dfrac{1}{2n})^{\frac{4n}{2n - 1}} - (\dfrac{2n}{2n + 1})(\dfrac{1}{2n})^{\frac{2n + 1}{2n - 1}} =

( 1 2 n ) 2 n + 1 2 n − 1 ( ( 4 n − 1 2 ) ( 1 2 n ) − 2 n 2 n + 1 ) = (\dfrac{1}{2n})^{\frac{2n + 1}{2n - 1}}((\dfrac{4n - 1}{2})(\dfrac{1}{2n}) - \dfrac{2n}{2n + 1}) =

1 2 ( 1 2 n ) 4 n 2 n − 1 ( 2 n − 1 2 n + 1 ) \boxed{\dfrac{1}{2}(\dfrac{1}{2n})^{\frac{4n}{2n - 1}}(\dfrac{2n - 1}{2n + 1})}

and

A 2 ( n ) = ∫ − ( 1 2 n ) 1 2 n − 1 0 ( x + ( 2 n − 1 ) j ( n ) + x 2 n ) d x = A_{2}(n) = \displaystyle\int_{-(\dfrac{1}{2n})^{\frac{1}{2n - 1}}}^{0} (x + (2n - 1)j(n) + x^{2n}) \:\ dx =

x 2 2 + ( 2 n − 1 ) j ( n ) x + 1 2 n + 1 x 2 n + 1 ∣ − ( 1 2 n ) 1 2 n − 1 0 = \dfrac{x^2}{2} + (2n - 1)j(n)x + \dfrac{1}{2n + 1}x^{2n + 1}|_{-(\dfrac{1}{2n})^{\frac{1}{2n - 1}}}^{0} =

− 1 2 ( 1 2 n ) 2 2 n − 1 + ( 2 n − 1 ) ( 1 2 n ) 2 n + 1 2 n − 1 + ( 1 2 n + 1 ) ( 1 2 n ) 2 n + 1 2 n − 1 = -\dfrac{1}{2}(\dfrac{1}{2n})^{\frac{2}{2n - 1}} + (2n - 1)(\dfrac{1}{2n})^{\frac{2n + 1}{2n - 1}} + (\dfrac{1}{2n + 1})(\dfrac{1}{2n})^{\frac{2n + 1}{2n - 1}} =

( 1 2 n ) 2 2 n − 1 ( − 1 2 + ( 2 n − 1 ) ( 1 2 n ) + ( 1 2 n + 1 ) ( 1 2 n ) ) = (\dfrac{1}{2n})^{\frac{2}{2n - 1}}(-\dfrac{1}{2} + (2n - 1)(\dfrac{1}{2n}) + (\dfrac{1}{2n + 1})(\dfrac{1}{2n})) =

1 2 ( 1 2 n ) 2 2 n − 1 ( 2 n − 1 2 n + 1 ) \boxed{\dfrac{1}{2}(\dfrac{1}{2n})^{\frac{2}{2n - 1}}(\dfrac{2n - 1}{2n + 1})}

⟹ A ( n ) = A 1 ( n ) + A 2 ( n ) = ( 4 n 2 + 1 ) ( 2 n − 1 ) 2 ( 2 n + 1 ) ( 1 2 n ) 4 n 2 n − 1 = ( 4 n 2 + 1 ) ( 2 n − 1 ) 2 ( 2 n + 1 ) ( 1 4 n 2 ) 2 n 2 n − 1 \implies A(n) = A_{1}(n) + A_{2}(n) = \dfrac{(4n^2 + 1)(2n - 1)}{2(2n + 1)}(\dfrac{1}{2n})^{\frac{4n}{2n - 1}} = \boxed{\dfrac{(4n^2 + 1)(2n - 1)}{2(2n + 1)}(\dfrac{1}{4n^2})^{\frac{2n}{2n - 1}}}

lim ⁡ n → ∞ A ( n ) = 1 2 lim ⁡ n → ∞ 2 n − 1 2 n + 1 ∗ \lim_{n \rightarrow \infty} A(n) = \dfrac{1}{2}\lim_{n \rightarrow \infty} \dfrac{2n - 1}{2n + 1} * lim ⁡ n → ∞ 4 n 2 + 1 ( 4 n 2 ) ( 4 n 2 ) 1 2 n − 1 = \lim_{n \rightarrow \infty} \dfrac{4n^2 + 1}{(4n^2)(4n^2)^{\frac{1}{2n - 1}}} =

1 2 lim ⁡ n → ∞ ( 1 + 1 4 n 2 ) ∗ lim ⁡ n → ∞ ( 1 4 n 2 ) 1 2 n − 1 = 1 2 lim ⁡ n → ∞ ( 1 4 n 2 ) 1 2 n − 1 \dfrac{1}{2}\lim_{n \rightarrow \infty} (1 + \dfrac{1}{4n^2}) * \lim_{n \rightarrow \infty} (\dfrac{1}{4n^2})^{\frac{1}{2n - 1}} = \dfrac{1}{2}\lim_{n \rightarrow \infty} (\dfrac{1}{4n^2})^{\frac{1}{2n - 1}}

and

lim ⁡ n → ∞ ( 1 4 n 2 ) 1 2 n − 1 = \lim_{n \rightarrow \infty} (\dfrac{1}{4n^2})^{\frac{1}{2n - 1}} = lim ⁡ n → ∞ e ( ln ⁡ ( 1 4 n 2 ) 2 n − 1 ) = \lim_{n \rightarrow \infty} e^{(\dfrac{\ln(\frac{1}{4n^2})}{2n - 1})} =

e lim ⁡ n → ∞ ( ln ⁡ ( 1 4 n 2 ) 2 n − 1 ) = e^{\lim_{n \rightarrow \infty} (\dfrac{\ln(\frac{1}{4n^2})}{2n - 1})} = e lim ⁡ n → ∞ ( − 1 n ) = e 0 = 1 e^{\lim_{n \rightarrow \infty} (-\dfrac{1}{n})} = e^{0} = 1

⟹ lim ⁡ n → ∞ A ( n ) = 1 2 \implies \boxed{\lim_{n \rightarrow \infty} A(n) = \dfrac{1}{2}}

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