(picture of text: https://i.stack.imgur.com/dSmPo.jpg)

$$\lim_{n \to \infty} \frac{2^{n+1} + n + 1}{(n+2)(2^n + n)}$$

My attempt:

$$\lim_{n \to \infty} \frac{2^{n+1}}{(n+2)(2^n + n)} + \lim_{n \to \infty} \frac{n}{(n+2)(2^n + n)} + \lim_{n \to \infty} \frac{1}{(n+2)(2^n + n)}$$

$$\lim_{n \to \infty} \frac{2^{n+1}}{(n+2)(2^n + n)} + 0$$

$$\lim_{n \to \infty} \frac{2}{(\frac{n}{2^n} + \frac{1}{2^{n-1}})(1 + \frac{n}{2^n})}$$

$$\lim_{n \to \infty} \frac{2}{0} = \infty$$

I split this limit into 3 parts and applied L'hospital rule to the second limit. The answer is wrong for some reason and l think it is the first limit which is messing things up.Any tips ?


Yes the problem is in the first limit. Note that there is no need to split and we may rewrite the whole term as $$\frac{2^{n+1}\left(1+\frac{n+1}{2^{n+1}}\right)}{n2^n(1+\frac{2}{n})(1+\frac{n}{2^{n}})}.$$ Can you take it from here?

  • $\begingroup$ Thanks, it evaluates to 0. $\endgroup$ – Aristotle Stagiritis Jan 15 at 8:28
  • 1
    $\begingroup$ Yes, you are correct! $\endgroup$ – Robert Z Jan 15 at 8:28

Your second-to-last step is wrong. You've divided twice by $2^n$!

$\frac{2}{a b}$ is not $\frac{1}{\frac{a}{2} \frac{b}{2}}$.

  • $\begingroup$ Yes, l notice that. In that case, the approach will fail as it will become 0/0. $\endgroup$ – Aristotle Stagiritis Jan 15 at 8:38

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