Menu Close

l-n-im-n-1-n-n-1-n-




Question Number 70360 by Hassen_Timol last updated on 03/Oct/19
l_(n→+∞) im    (((√(n + 1 ))− n)/( (√(n + 1)) + n))  =  ?
$$\underset{{n}\rightarrow+\infty} {\mathrm{l}im}\:\:\:\:\frac{\sqrt{{n}\:+\:\mathrm{1}\:}−\:{n}}{\:\sqrt{{n}\:+\:\mathrm{1}}\:+\:{n}}\:\:=\:\:? \\ $$
Answered by mind is power last updated on 03/Oct/19
(((√(n+1))−n)/( (√(n+1))+n))=((n((√((n+1)/n^2 ))−1))/(n((√((n+1)/n^2 ))+1)))=(((√((1/n)+(1/n^2 )))−1)/( (√((1/n)+(1/n^2 )))+1))→−1
$$\frac{\sqrt{{n}+\mathrm{1}}−{n}}{\:\sqrt{{n}+\mathrm{1}}+{n}}=\frac{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}−\mathrm{1}\right)}{{n}\left(\sqrt{\frac{{n}+\mathrm{1}}{{n}^{\mathrm{2}} }}+\mathrm{1}\right)}=\frac{\sqrt{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}−\mathrm{1}}{\:\sqrt{\frac{\mathrm{1}}{{n}}+\frac{\mathrm{1}}{{n}^{\mathrm{2}} }}+\mathrm{1}}\rightarrow−\mathrm{1} \\ $$
Commented by Hassen_Timol last updated on 03/Oct/19
Thank you sooo much !!!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sooo}\:\mathrm{much}\:!!! \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *