Question

Limita unui șir de integrale va rog... ​

Answer 1

$\displaystyle a)\,\,\,I_n = \int_{1}^{\sqrt{2}}\sqrt[n]{x^n+x^{n+2}}\, dx =\int_{1}^{\sqrt{2}}\sqrt[n]{x^n(1+x^{2}})\, dx =\\ \\ = \int_{1}^{\sqrt{2}}x\sqrt[n]{(1+x^{2}})\, dx \\ \\\lim\limits_{n\to \infty} I_{n} = \lim\limits_{n \to \infty}\int_{1}^{\sqrt{2}}x\sqrt[n]{x^2}\, dx =\lim\limits_{n\to \infty}\int_{1}^{\sqrt{2}}x^{\frac{n+2}{n}}\,dx=\\ \\ = \int_{1}^{\sqrt{2}}\lim\limits_{n\to \infty}x^{\frac{n+2}{n}}\, dx = \int_{1}^{\sqrt{2}}x\,dx = \frac{x^2}{2}\Big|_{1}^{\sqrt{2}}=\\ \\ = 1-\dfrac{1}{2} =\boxed{\boxed{\frac{1}{2}}}$

$\\\displaystyle b)\,\,\,L =\lim\limits_{n\to \infty}n\left(I_n-\dfrac{1}{2}\right) =\lim\limits_{n\to \infty}n\left(\int_{1}^{\sqrt{2}}x\sqrt[n]{1+x^2}\, dx-\dfrac{1}{2}\right) =\\ \\ =\lim\limits_{n\to \infty}n\left(\frac{1}{2}\int_{1}^{\sqrt{2}}(1+x^2)'(1+x^2)^{\frac{1}{n}}\, dx-\dfrac{1}{2}\right) =\\ \\ =\lim\limits_{n\to \infty}n\left(\frac{(1+x^2)^{\frac{1}{n}+1}}{2\left(\frac{1}{n}+1\right)}\Bigg|_{1}^{\sqrt{2}}-\dfrac{1}{2}\right)\\ \\\text{Notez }\dfrac{1}{n} = m \\ \\ L =\lim\limits_{m\to 0}\dfrac{1}{m}\left(\dfrac{3^{m+1}-2^{m+1}}{2(m+1)}-\dfrac{1}{2}\right)$

$\text{Consider }f:\mathbb{R} \backslash\{-1;0\} \to \mathbb{R},\,\,f(x) = \dfrac{1}{x}\left(\dfrac{3^{x+1}-2^{x+1}}{2(x+1)}-\dfrac{1}{2}\right)\\ \\ L = \lim\limits_{x\to 0} f(x)\\\\ L=\lim\limits_{x\to 0}\dfrac{1}{x}\left(\dfrac{3^{x+1}-2^{x+1}}{2(x+1)}-\dfrac{1}{2}\right) =\\ \\ = \lim\limits_{x\to 0}\dfrac{2\cdot 3^{x+1}-2\cdot 2^{x+1}-2x-2}{4x^2+4x} \overset{L'H}{=} \\ \\ \overset{L'H}{=}\lim\limits_{x\to 0}\dfrac{3^{x+1}\cdot 2\ln 3-2^{x+1}\cdot 2\ln 2-2}{8x+4}=\dfrac{6\ln 3-4\ln 2-2}{4}=\\ \\ = \boxed{\boxed{\dfrac{3\ln 3-2\ln 2-1}{2}}}$