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I'm trying to use Mathematica to compute a limit, because I have no idea how to compute it by myself.

The limit should be: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}$$

Any ideas?

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  • $\begingroup$ Welcome to Mathematica Stack Exchange! Can you show anything you've tried? You might like to refer to the excellent documentation within Mathematica, for example on Limit. You can also use Wolfram|Alpha from within Mathematica by typing = = in a new cell and entering your query in brief English as you might on the web. $\endgroup$ Mar 24, 2015 at 15:06
  • $\begingroup$ Limit[(2^(n + 1) - n - 2)/2^n, n -> \[Infinity]] === 2. This becomes evident when you distribute the denominator and look at the behavior of each term. $\endgroup$ Mar 24, 2015 at 15:33

1 Answer 1

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Your limit $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}=2$$ may be decomposed into the sum: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}-n-2}{2^{n}}=\lim _{n\rightarrow \infty }\dfrac {2^{n+1}}{2^{n}}-\dfrac {n}{2^{n}}-\dfrac {2}{2^{n}}=2-0-0=2$$ And each limit: $$\lim _{n\rightarrow \infty }\dfrac {2^{n+1}}{2^{n}}=2$$ $$\lim _{n\rightarrow \infty }\dfrac {n}{2^{n}}=0$$ $$\lim _{n\rightarrow \infty }\dfrac {2}{2^{n}}=0$$

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