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Question Number 188060 by mathlove last updated on 25/Feb/23
Answered by aleks041103 last updated on 26/Feb/23
lim_(x→0) (Π_(k=2) ^n ((cos(kx)))^(1/k) )^(1/x^2 ) = =lim_(x→0) (Π_(k=2) ^n (1−((k^2 x^2 )/2)+o(x^2 ))^(1/k) )^(1/x^2 ) = =Π_(k=2) ^n lim_(x→0) (1−((kx^2 )/2)+o(x^2 )+o(((k^2 x^2 )/2)+o(x^2 )))^(1/x^2 ) = =Π_(k=2) ^n lim_(x→0) (1−((kx^2 )/2)+o(x^2 ))^(1/x^2 ) = =Π_(k=2) ^n e^(−k/2) =e^(−(1/2)Σ_(k=2) ^n k) =e^(−(1/2)(((n(n+1))/2)−1)) = =e^(−((n^2 +n−2)/4))
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\sqrt[{{k}}]{{cos}\left({kx}\right)}\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\left(\mathrm{1}−\frac{{k}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)^{\mathrm{1}/{k}} \right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\mathrm{1}−\frac{{kx}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)+{o}\left(\frac{{k}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\mathrm{1}−\frac{{kx}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{k}=\mathrm{2}} {\overset{{n}} {\prod}}{e}^{−{k}/\mathrm{2}} ={e}^{−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{2}} {\overset{{n}} {\sum}}{k}} ={e}^{−\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{n}\left({n}+\mathrm{1}\right)}{\mathrm{2}}−\mathrm{1}\right)} = \\ $$$$={e}^{−\frac{{n}^{\mathrm{2}} +{n}−\mathrm{2}}{\mathrm{4}}} \\ $$
Answered by aleks041103 last updated on 26/Feb/23
lim_(x→0) (Π_(k=1) ^n cos(kx))^(1/x^2 ) = =Π_(k=1) ^n lim_(x→0) (cos(kx))^(1/x^2 ) = =Π_(k=1) ^n lim_(x→0) (1−((k^2 x^2 )/2)+o(x^2 ))^(1/x^2 ) = =Π_(k=1) ^n e^(−(k^2 /2)) =exp(−(1/2)Σ_(k=1) ^n k^2 )= =exp(−(1/2) ((n(n+1)(2n+1))/6))= =e^(−((n(n+1)(2n+1))/(12)))
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{cos}\left({kx}\right)\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left({cos}\left({kx}\right)\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\mathrm{1}−\frac{{k}^{\mathrm{2}} {x}^{\mathrm{2}} }{\mathrm{2}}+{o}\left({x}^{\mathrm{2}} \right)\right)^{\mathrm{1}/{x}^{\mathrm{2}} } = \\ $$$$=\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{e}^{−\frac{{k}^{\mathrm{2}} }{\mathrm{2}}} ={exp}\left(−\frac{\mathrm{1}}{\mathrm{2}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}^{\mathrm{2}} \right)= \\ $$$$={exp}\left(−\frac{\mathrm{1}}{\mathrm{2}}\:\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{6}}\right)= \\ $$$$={e}^{−\frac{{n}\left({n}+\mathrm{1}\right)\left(\mathrm{2}{n}+\mathrm{1}\right)}{\mathrm{12}}} \\ $$
Answered by aleks041103 last updated on 26/Feb/23
lim_(x→0) ((1/n)Σ_(k=1) ^n e^(kx) )^(1/x) = =lim_(x→0) ((1/n)Σ_(k=1) ^n (1+kx+o(x)))^(1/x) = =lim_(x→0) ((1/n)(n+(Σ_(k=1) ^n k)x+o(x)))^(1/x) = =lim_(x→0) (1+((n+1)/2)x+o(x))^(1/x) = =e^((n+1)/2)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{kx}} \right)^{\mathrm{1}/{x}} = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\mathrm{1}+{kx}+{o}\left({x}\right)\right)\right)^{\mathrm{1}/{x}} = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}}{{n}}\left({n}+\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{k}\right){x}+{o}\left({x}\right)\right)\right)^{\mathrm{1}/{x}} = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\mathrm{1}+\frac{{n}+\mathrm{1}}{\mathrm{2}}{x}+{o}\left({x}\right)\right)^{\mathrm{1}/{x}} = \\ $$$$={e}^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \\ $$
Answered by aleks041103 last updated on 26/Feb/23
lim_(x→0) ((1/n)Σ_(k=1) ^n a_k ^x )^(1/x) = =lim_(x→0) ((1/n)Σ_(k=1) ^n e^(ln(a_k )x) )^(1/x) = =lim_(x→0) (1+ln(((a_1 a_2 ...a_n ))^(1/n) )x+o(x))^(1/x) = =e^(ln(((a_1 ...a_n ))^(1/n) )) =((Π_(k=1) ^n a_k ))^(1/n)
$$\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{k}} ^{{x}} \right)^{\mathrm{1}/{x}} = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\frac{\mathrm{1}}{{n}}\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{e}^{{ln}\left({a}_{{k}} \right){x}} \right)^{\mathrm{1}/{x}} = \\ $$$$=\underset{{x}\rightarrow\mathrm{0}} {{lim}}\left(\mathrm{1}+{ln}\left(\sqrt[{{n}}]{{a}_{\mathrm{1}} {a}_{\mathrm{2}} …{a}_{{n}} }\right){x}+{o}\left({x}\right)\right)^{\mathrm{1}/{x}} = \\ $$$$={e}^{{ln}\left(\sqrt[{{n}}]{{a}_{\mathrm{1}} …{a}_{{n}} }\right)} =\sqrt[{{n}}]{\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}{a}_{{k}} } \\ $$

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