Question #191550

Find the following limits


lim( sqrt2n+1− Sqrt2n)

n→∞


lim3n^3 −n+8/(4n(n−1)(n−2))

n→∞



b) Consider the sequence (an) = (−1)^n − 2n

 Explain whether the sequence is monotone increasing or decreasing, whether it is monotone and if limn→∞(an) = −∞



1
Expert's answer
2021-05-13T08:37:30-0400

(a) (i)limn(2n+12n)lim_{n\to \infty} (\sqrt{2n+1}-\sqrt{2n})


          Rationalising the given expression-

                

        =limn(2n+12n)×2n+1+2n2n+1+2n=limn(2n+1)2(2n)22n+1+2n=limn12n+1+2n= lim_{n\to \infty} (\sqrt{2n+1}-\sqrt{2n})\times \dfrac{ \sqrt{2n+1}+\sqrt{2n}}{\sqrt{2n+1}+\sqrt{2n}} \\[9pt] = lim_{n\to \infty} \dfrac{ (\sqrt{2n+1})^2-(\sqrt{2n})^2}{\sqrt{2n+1}+\sqrt{2n}} \\[9pt] =lim _{n\to \infty} \dfrac{1}{\sqrt{2n+1}+\sqrt{2n}}

        Applying limits


         = 0


(ii) limn3n3n+84n(n1)(n2)lim_{n\to \infty} \dfrac{3n^3-n+8}{4n(n-1)(n-2)}

 

    Now writing it my simplifying-

     

   =limn3n3n+84n(n22nn+2)=limn3n3n+84n312n2+8n=limnn3(31n2+8n3)n3(412n+8n2)=30+040+0=34=lim_{n\to \infty} \dfrac{3n^3-n+8}{4n(n^2-2n-n+2)} \\[9pt] =lim_{n\to \infty} \dfrac{3n^3-n+8}{4n^3-12n^2+8n} \\[9pt]=lim_{n\to \infty} \dfrac{n^3(3-\frac{1}{n^2}+\frac{8}{n^3})}{n^3(4-\frac{12}{n}+\frac{8}{n^2})} \\[9pt]=\dfrac{3-0+0}{4-0+0}=\dfrac{3}{4}

    


(b) Given, sequence

        an=(1)n2na_n=(-1)^n-2n

        a1=12=3a2=14=3a3=16=7a4=7a5=110=11a6=112=11a_1=-1-2=-3\\ a_2=1-4=-3\\ a_3=-1-6=-7\\ a_4=-7\\ a_5=-1-10=-11\\ a_6=1-12=-11\\


<an>=3,3,7,7,11,11...\Rightarrow <a_n>=-3,-3,-7,-7,-11,-11...


 Here, a1a2a3a4a5a6...a_1\ge a_2\ge a_3\ge a_4\ge a_5\ge a_6...


we know a sequence <an><a_n> is monotonic decreasing if an+1annNa_{n+1}\le a_n \forall n\in N


Hence Given sequence is monotone decreasing.

So limn(an)=.lim_{n\to \infty}(a_n)=-\infty.


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