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Can anyone solve the following two for me:

1-Let $a$, $b$, $c$ and $d$ be real numbers satisfying $a>b$ and $c>d$. Does this imply that $ac>bd$? Prove or disprove.

2-From analysis and the concepts of limits

find $$ \lim_{n\to\infty}\frac{3^n+n^3}{(2n)^2+2^{2n}}\,. $$

I really appreciate your efforts.

Thank you.

LoveMath
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1 Answers1

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1- take $a=2, b=1$ and $c=-2, d=-3$ and you find a counterexample, yet the result is true if the reals $a,b,c,d$ are positive.

2- we know that $n^3=_\infty o(3^n)$ and $(2n)^2=_\infty o(2^{2n})$ so $$\lim_{n\to\infty}\frac{3^n+n^3}{(2n)^2+2^{2n}}=\lim_{n\to\infty}\frac{3^n}{2^{2n}}=\lim_{n\to\infty}\left(\frac{3}{4}\right)^n=0$$

  • Thanks Sami, but can you make it more clear please? I mean what do you mean by $n^3=\infty o(3^n)$ and $(2n)^2=\infty o(2^{2n})$ and how $$\lim_{n\to\infty}\frac{3^n+n^3}{(2n)^2+2^{2n}} equals \lim_{n\to\infty}\frac{3^n}{2^{2n}}=\lim_{n\to\infty}\left(\frac{3}{4}\right)^n=0$$ – LoveMath Apr 12 '13 at 08:14
  • @user50382 Simply that means $\lim_{n\to\infty}\frac{n^3}{3^n}=0$ and $\lim_{n\to\infty}\frac{(2n)^2}{2^{2n}}=0$ and then you can write $\frac{3^n+n^3}{(2n)^2+2^{2n}}=\frac{3^n}{2^{2n}}\frac{1+\frac{n^3}{3^n}}{1+\frac{(2n)^2}{2^{2n}}}$ and then you pass to limit. –  Apr 12 '13 at 08:20
  • Wow..I got confused, now I am going to ask you why \lim_{n\to\infty}\frac{(2n)^2}{2^{2n}}=0 and why \lim_{n\to\infty}\frac{n^3}{3^n}=0 – LoveMath Apr 12 '13 at 08:26
  • If you know the elementary result $\lim_{n\to\infty}\frac{\ln n}{n}=0$ then you can see that $\lim \frac{(2n)^2}{2^{2n}}=0$ by taking the $\ln(\frac{(2n)^2}{2^{2n}})$ –  Apr 12 '13 at 08:31