Question Number 216144 by klipto last updated on 28/Jan/25
![1. Lim_(n→∞) [(1/n^2 )+(2/n^2 )+(3/n^2 )+...+((n+1)/n^2 )] 2. lim_(x→0) (((3sin5x)/x))^((1−cos4x)/x^2 )](https://www.tinkutara.com/question/Q216144.png)
$$\mathrm{1}.\:\boldsymbol{\mathrm{Lim}}_{\mathrm{n}\rightarrow\infty} \left[\frac{\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }+\frac{\mathrm{2}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }+\frac{\mathrm{3}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }+…+\frac{\boldsymbol{\mathrm{n}}+\mathrm{1}}{\boldsymbol{\mathrm{n}}^{\mathrm{2}} }\right] \\ $$$$\mathrm{2}.\:\boldsymbol{\mathrm{lim}}_{\boldsymbol{\mathrm{x}}\rightarrow\mathrm{0}} \left(\frac{\mathrm{3}\boldsymbol{\mathrm{sin}}\mathrm{5}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}}\right)^{\frac{\mathrm{1}−\boldsymbol{\mathrm{cos}}\mathrm{4}\boldsymbol{\mathrm{x}}}{\boldsymbol{\mathrm{x}}^{\mathrm{2}} }} \\ $$
Answered by Ghisom last updated on 28/Jan/25
$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\:\frac{{k}}{{n}^{\mathrm{2}} }\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\:{k}}{{n}^{\mathrm{2}} }\:=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\:\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}{\mathrm{2}{n}^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Answered by mr W last updated on 28/Jan/25
$$\left.\mathrm{1}\right) \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\underset{{k}=\mathrm{1}} {\overset{{n}+\mathrm{1}} {\sum}}\frac{{k}}{{n}^{\mathrm{2}} } \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {xdx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$
Commented by klipto last updated on 28/Jan/25
$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{bossman}} \\ $$
Answered by mr W last updated on 28/Jan/25
$$\left.\mathrm{2}\right) \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{15}\:\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{5}{x}}\right)^{\frac{\mathrm{2}\:\mathrm{sin}^{\mathrm{2}} \:\mathrm{2}{x}}{{x}^{\mathrm{2}} }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\frac{\mathrm{15}\:\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{5}{x}}\right)^{\mathrm{8}\left(\frac{\mathrm{sin}\:\mathrm{2}{x}}{\mathrm{2}{x}}\right)^{\mathrm{2}} } \\ $$$$=\left(\mathrm{15}×\mathrm{1}\right)^{\mathrm{8}×\mathrm{1}^{\mathrm{2}} } \\ $$$$=\mathrm{15}^{\mathrm{8}} \\ $$
Commented by klipto last updated on 28/Jan/25
$$\boldsymbol{\mathrm{thanks}}\:\boldsymbol{\mathrm{ghisom}} \\ $$
Commented by Ghisom last updated on 28/Jan/25
$$\mathrm{yes}.\:\mathrm{I}\:\mathrm{missed}\:\mathrm{the}\:\mathrm{power}… \\ $$
Answered by MrGaster last updated on 29/Jan/25
![(1):=lim_(n→∞) [((1+2+3+…+(n+1))/n^2 )] =lim_(n→∞) [(((n+1)(n+2))/(2/n^2 ))] =lim_(n→∞) [((n^2 +3n+2)/(2n^2 ))] =(1/2) (2):=lim_(x→0) (3∙((sin 5x)/(5x))∙5)^((1−(1−(((4x^2 ))/(2!))+…))/x^2 ) =lim_(x→0) (15∙((sin 5x)/(5x)))^((8x^2 )/x^2 ) =lim_(x→0) 15^8 (lim_(x→0) ((sin 5x)/(5x))=1) =15^8](https://www.tinkutara.com/question/Q216165.png)
$$\left(\mathrm{1}\right):=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\mathrm{1}+\mathrm{2}+\mathrm{3}+\ldots+\left({n}+\mathrm{1}\right)}{{n}^{\mathrm{2}} }\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{\left({n}+\mathrm{1}\right)\left({n}+\mathrm{2}\right)}{\frac{\mathrm{2}}{{n}^{\mathrm{2}} }}\right] \\ $$$$=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left[\frac{{n}^{\mathrm{2}} +\mathrm{3}{n}+\mathrm{2}}{\mathrm{2}{n}^{\mathrm{2}} }\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\mathrm{2}\right):=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{3}\centerdot\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{5}{x}}\centerdot\mathrm{5}\right)^{\frac{\mathrm{1}−\left(\mathrm{1}−\frac{\left(\mathrm{4}{x}^{\mathrm{2}} \right)}{\mathrm{2}!}+\ldots\right)}{{x}^{\mathrm{2}} }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\left(\mathrm{15}\centerdot\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{5}{x}}\right)^{\frac{\mathrm{8}{x}^{\mathrm{2}} }{{x}^{\mathrm{2}} }} \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}15}^{\mathrm{8}} \left(\underset{{x}\rightarrow\mathrm{0}} {\mathrm{lim}}\frac{\mathrm{sin}\:\mathrm{5}{x}}{\mathrm{5}{x}}=\mathrm{1}\right) \\ $$$$=\mathrm{15}^{\mathrm{8}} \\ $$