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Just when i thought i finally got the hang of limits, i stumbled upon this:

$$\frac{n^4+4^n}{n+4^{n+1}}$$

Now, this kinda makes sense in my head because $4^n$ grows a lot faster than $n^4$, let alone $n$. Now my question is if this assumption is a valid tool for solving this limit i.e. if i divide both denominator and numerator with $4^n$, can that be user without further explanation or proof.

This, of course, gives the result of $\frac{1}{4}$ which seems to be correct.

EDIT: $n \in \mathbb{N}$

Transcendental
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    Yes, you can do that. Or use L'hopital's rule 5 times assuming $n \in \Bbb R$, the limit is the same if $n \in \Bbb N$ as long as $n \rightarrow \infty$. –  Feb 08 '15 at 19:41
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    You kinda forgot to specify $\lim\limits_{n\to?}$ – barak manos Feb 08 '15 at 19:42

4 Answers4

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Note that:

$$\frac{n^4+4^n}{n+4^{n+1}}=\frac{4^{-n}}{4^{-n}}\frac{n^4+4^n}{n+4^{n+1}}=\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^{n}}+4}$$

Now you have:

$$\frac{n^4}{4^n} \to 0$$

$$\frac{n}{4^n} \to 0$$

when $n \to \infty$, so:

$$\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^{n}}+4} \to \frac{0+1}{0+4}=\frac{1}{4}$$

agha
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  • I think you need to show that exponential functions grow quicker than polynomials. (Or is that given?) – Columbo Feb 08 '15 at 19:46
  • I kinda wrote what you just said.... That's the question, if i can use that. It's what @Columbo said, is that given or do i have to show it. – Transcendental Feb 08 '15 at 19:48
  • @Transcendental It depends. If you have just started learning about limits at school you should do it and additionally it will help in the future. Otherwise it is a well-known fact and you can use it as given. – Stefan4024 Feb 08 '15 at 20:14
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Hint: $$ \frac{n^4+4^n}{n+4^{n+1}}=\frac{\frac{n^4}{4^n}+1}{\frac{n}{4^n}+4} $$

Joe
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As you said you can do $$\lim_{n\rightarrow\infty} \frac{n^4+4^n}{n+4^{n+1}} = \lim_{n\rightarrow\infty} \frac{\frac{n^4}{4^n}+1}{\frac{n}{4^n}+4} = \frac{\lim_{n\rightarrow\infty} \frac{n^4}{4^n}+1}{\lim_{n\rightarrow\infty} \frac{n}{4^n}+4} = \frac{0+1}{0+4}$$

But it's necessary that you already have proven in you lecture, that

  • $\lim_{n\rightarrow\infty}\frac{n^k}{q^n} = 0$ when $|q| > 1$ and $k\in\mathbb N$
  • $\lim_{n\rightarrow\infty}\frac{a_n}{b_n} = \frac{\lim_{n\rightarrow\infty} a_n}{\lim_{n\rightarrow\infty}b_n}$ when $(a_n)$ and $(b_n)$ are convergent
  • $\lim_{n\rightarrow\infty}(a_n+b_n) = \lim_{n\rightarrow\infty} a_n+\lim_{n\rightarrow\infty}b_n$ when $(a_n)$ and $(b_n)$ are convergent

I guess you will find similar theorems in your script...

Stephan Kulla
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  • Why must $k \in \mathbb{N}$? – Columbo Feb 08 '15 at 19:52
  • @Columbo: It doesn't need to, but in the case of the OP one time we have $k=1$ and one time $k=4$. That's why I took $k\in\mathbb N$... I also think this theorem is often proven only for the case $k\in\mathbb N$ because the general exponentiation is mostly introduced after the limit... – Stephan Kulla Feb 08 '15 at 19:57
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exponential function $4^n$ is greater than any power of $n$ for $n$ large. that is $\lim_{n \to \infty}\dfrac{n^k}{4^n} = 0$ for any $k.$ assuming this,

$\dfrac{n^4 + 4^n}{n + 4^{n+1}} = \dfrac{ 4^n + \cdots}{4^{n+1} + \cdots} = \dfrac{1}{4} + \cdots$

establishing the limit. suppose $\lim_{n \to \infty}\dfrac{n^k}{4^n} = L$ replacing $n$ by $2n$ we shoud get the same limit. therefore $$L = \lim_{n \to \infty}\dfrac{n^k}{4^n} = \lim_{n \to \infty}\dfrac{(2n)^k}{4^{2n}} = \lim_{n \to \infty}\frac{2^k}{4^n}\dfrac{n^k}{4^n}= \lim_{n \to \infty}\frac{2^k}{4^n}L = 0.$$

abel
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