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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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3  
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
1  
No, it converges to 5. –  David Mitra Apr 2 '12 at 18:34
1  
How do you know when it diverges and when it converges? –  Jeremy Apr 2 '12 at 18:35

1 Answer 1

up vote 3 down vote accepted

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.)

Though somewhat badly stated, your reasoning is correct.

Divide everything by the "dominant term" (not the least dominant term). This is the factorial here:

$$\tag{1} { {5n!} +{n^9}-{2^n} \over {2^n}+{n!}+ n^{100} } ={ 5+{n^9\over n!}-{2^n\over n!} \over 1+{2^n\over n!} + {n^{100}\over n!}} $$ 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.

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Thanks! +1... :-) –  Aryabhata Apr 2 '12 at 20:12

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