if zn=(an+bn)1/nz_n=(a^n+b^n)^{1/n}zn=(an+bn)1/n where 0<a<b0<a<b0<a<b , then
limn→∞zn=b\displaystyle{\lim_{n\to \infin}}z_n=bn→∞limzn=b
(an+bn)1/n=b((a/b)n+1)1/n(a^n+b^n)^{1/n}=b((a/b)^n+1)^{1/n}(an+bn)1/n=b((a/b)n+1)1/n
limn→∞(a/b)n=0\displaystyle{\lim_{n\to \infin}}(a/b)^n=0n→∞lim(a/b)n=0
limn→∞((a/b)n+1)1/n=0\displaystyle{\lim_{n\to \infin}}((a/b)^n+1)^{1/n}=0n→∞lim((a/b)n+1)1/n=0
limn→∞(b((a/b)n+1)1/n)=b⋅1=b\displaystyle{\lim_{n\to \infin}}(b((a/b)^n+1)^{1/n})=b\cdot1=bn→∞lim(b((a/b)n+1)1/n)=b⋅1=b
Need a fast expert's response?
and get a quick answer at the best price
for any assignment or question with DETAILED EXPLANATIONS!
Comments
Leave a comment