# Finding the limits of a series of sequences

I need to find the limits of the following sequences

$$\frac{({5n!}+{n^9}-{2^n})}{({2^n}+{n!}+{n^{100}})}$$

So far I've divided by the least dominant term, which I think is ${n!}$ and that pretty much cancels the whole thing down to 5. Is that my limit or am I horrifically wrong?

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That's exactly right. –  Robert Israel Apr 2 '12 at 18:22
If you are not certain, I would recommend to look for squeezing sequences. –  AD. Apr 2 '12 at 18:24
its pretty horrible to do the squeeze rule on that. If the limit is 5 then that means it diverges right? –  Jeremy Apr 2 '12 at 18:25
No, it converges to 5. –  David Mitra Apr 2 '12 at 18:34
How do you know when it diverges and when it converges? –  Jeremy Apr 2 '12 at 18:35

I will assume you want to evaluate $$\lim_{n\rightarrow\infty} { {5n!} +{n^9}-{2^n} \over {2^n}+{n!}+ n^{100} } .$$ (This is just one sequence, of course.)

$$\tag{1} { {5n!} +{n^9}-{2^n} \over {2^n}+{n!}+ n^{100} } ={ 5+{n^9\over n!}-{2^n\over n!} \over 1+{2^n\over n!} + {n^{100}\over n!}}$$ The whole thing does not quite "cancel down to 5". However, this is true as far as taking the limit as $n$ tends to infinity is concerned. Indeed, you know (hopefully) that $$\lim_{n\rightarrow\infty} {2^n\over n!} =0$$ and that for any real number $a$ $$\lim_{n\rightarrow\infty} {n^a\over n!}=0.$$ So, using these facts and $(1)$, we have \eqalign{ \lim_{n\rightarrow\infty} { {5n!} +{n^9}-{2^n} \over {2^n}+{n!}+ n^{100} } &=\lim_{n\rightarrow\infty}{ 5+{n^9\over n!}-{2^n\over n!} \over 1+{2^n\over n!} + {n^{100}\over n!} }\cr &=\lim_{n\rightarrow\infty}{ 5+0+0\over 1+0+0}\cr&=5. }
From your comments concerning convergence and divergence, recall the definitions: the sequence $(a_n)$ converges if $\lim\limits_{n\rightarrow\infty} a_n$ exists. In this case we say the sequence converges to the value of the limit. The sequence $(a_n)$ diverges if $\lim\limits_{n\rightarrow\infty} a_n$ does not exist.
Informally a sequence converges if its terms get closer and closer to some number $L$ as $n$ gets larger and larger, and then we say the sequence converges to $L$. If the terms do not "settle down" to any number, the sequence diverges.
For your sequence, as $n$ gets larger and larger, the non constant bits of the right hand side of $(1)$ get closer and closer to 0. Then, the entire term on the right hand side of $(1)$ gets closer and closer to 5. So the sequence converges to 5.