Salut,
[tex]Folosim\ formula:\ a^2-b^2=(a-b)\cdot(a+b).\\\\n+6-(n-1)=\sqrt{(n+6)^2}-\sqrt{(n-1)^2},\ sau\\\\n+6-n+1=(\sqrt{n+6}-\sqrt{n-1})\cdot(\sqrt{n+6}+\sqrt{n-1}),\ sau\\\\7=(\sqrt{n+6}-\sqrt{n-1})\cdot(\sqrt{n+6}+\sqrt{n-1})\Longrightarrow\\\\\Longrightarrow\sqrt{n+6}-\sqrt{n-1}=\dfrac{7}{\sqrt{n+6}+\sqrt{n-1}}\ (1).\\\\n^2+1-(n^2+3)=\sqrt{(n^2+1)^2}-\sqrt{(n^2+3)^2},\ sau\\\\n^2+1-n^2-3=(\sqrt{n^2+1}-\sqrt{n^2+3})\cdot(\sqrt{n^2+1}+\sqrt{n^2+3}),\ sau\\\\-2=(\sqrt{n^2+1}-\sqrt{n^2+3})\cdot(\sqrt{n^2+1}+\sqrt{n^2+3})\Longrightarrow\\\\\Longrightarrow\sqrt{n^2+1}-\sqrt{n^2+3}=\dfrac{-2}{\sqrt{n^2+1}+\sqrt{n^2+3}}\ (2).\\\\\hat{I}mp\breve{a}r\c{t}im\ rela\c{t}ia\ (1)\ la\ rela\c{t}ia\ (2),\ membru\ cu\ membru:\\\\\dfrac{\sqrt{n+6}-\sqrt{n-1}}{\sqrt{n^2+1}-\sqrt{n^2+3}}=-\dfrac{7}2\cdot\dfrac{\sqrt{n^2+1}+\sqrt{n^2+3}}{\sqrt{n+6}+\sqrt{n-1}}=-\dfrac{7}2\cdot\dfrac{n\cdot\left(\sqrt{1+\dfrac{1}{n^2}}+\sqrt{1+\dfrac{3}{n^2}}\right)}{\sqrt n\cdot\left(\sqrt{1+\dfrac{6}{n}}+\sqrt{1-\dfrac{1}{n}}\right)}.[/tex]
Scriem fracția așa:
[tex]-\dfrac{7\sqrt n}2\cdot\dfrac{\sqrt{1+\dfrac{1}{n^2}}+\sqrt{1+\dfrac{3}{n^2}}}{\sqrt{1+\dfrac{6}{n}}+\sqrt{1-\dfrac{1}{n}}}.[/tex]
Când trecem la limită, pentru n care tinde la +∞, a doua fracție tinde la:
[tex]\dfrac{\sqrt{1+0}+\sqrt{1+0}}{\sqrt{1+0}+\sqrt{1-0}}=\dfrac{1+1}{1+1}=1.[/tex]
Prima fracție tinde la --∞ (minus infinit), deci valoarea limitei este:
L = --∞.
Ai înțeles rezolvarea ?
Green eyes.
