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$$\lim_{n\to\infty}\frac{5\cdot 2^n-4}{2^n-1}=5.$$ I have already proved this just using the definition of convergence. How do I go about proving this only using the sandwich theorem and sum/product/quotient rule?

I can divide the whole expression by $2^n$ and say the following: I know $5 - \frac4{2^n}$ tends to $5$ as $n\to\infty$. And I also know that $1 -\frac1{2^n}$ tends to $1$ as $n\to\infty$ so using the quotient rule, I can say that the whole expression tends to $\frac51=5$ as $n\to\infty$. At first glance this seems correct to me. Is there a more sophisticated way of proving this?

Vladimir Nabokov
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1 Answers1

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That's absolutely correct. Since for all $n$ we have $$\frac{5\cdot 2^n-4}{2^n+1}=\cfrac{5-\frac4{2^n}}{1+\frac1{2^n}},$$ where the numerator and denominator of the RHS expression converge (with the denominator converging to a non-$0$ number), then the limit of quotients is the quotient of limits, just as you've done.

Cameron Buie
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  • Should I prove before hand how $1/2^n$ converges to 0? – Vladimir Nabokov Oct 23 '13 at 13:52
  • That would certainly be the simplest way to show that the numerator of the RHS expression tends to $5$ and the denominator tends to $1$ (again, using the rule about quotients). It's a fairly easy proof, too. Just show that $n\le 2^n$ for all $n$ by induction, then apply Archimedean property. Or you can just use a direct $\epsilon$-$N$ argument, if you've proven the existence of logarithms. – Cameron Buie Oct 23 '13 at 14:00