Suppose a1=2,an+1=12(an+2an). assuming an has a limit, find limn→∞an= . hint: let a=limn→∞. then, since an+1=12(an+2an), we have a=12(a+2a). now solve for
a.

Looks like the sequence might be

[tex]a_{n+1}=\dfrac{a_n+\frac2{a_n}}2[/tex]

Assume that [tex]\displaystyle\lim_{n\to\infty}a_n=L[/tex]. Then taking the limit of both sides, we have

[tex]L=\dfrac{L+\frac2L}2\implies L=\pm\sqrt2[/tex]

But in order for the sequence to converge, only one of these values must be true. Since [tex]a(1)=2>0[/tex], each successive number will also be positive, so the limit must be [tex]L=\sqrt2[/tex].

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Suppose a1=2,an+1=12(an+2an). assuming an has a limit, find limn→∞an= . hint: let a=limn→∞. then, since an+1=12(an+2an), we have a=12(a+2a). now solve for a.

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Suppose a1=2,an+1=12(an+2an). assuming an has a limit, find limn→∞an= . hint: let a=limn→∞. then, since an+1=12(an+2an), we have a=12(a+2a). now solve for
a.