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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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    That's exactly right.2012-04-02
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    If you are not certain, I would recommend to look for squeezing sequences.2012-04-02
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    its pretty horrible to do the squeeze rule on that. If the limit is 5 then that means it diverges right?2012-04-02
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    No, it converges to 5.2012-04-02
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    How do you know when it diverges and when it converges?2012-04-02
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    By squeezing sequences I mean that you should (1) try to guess which term is dominant and (2) estimate estimate each individual term with respect to the dominant term from above and below, and finally (3) gather the estimate to yield an estimate of your expression (from above and below).2012-04-02
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    If the limit as $n$ tends to infinity exists, the sequence converges to that limit value. If the limit doesn't exist (including the case where it is infinite), the sequence diverges.2012-04-02
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    I think I get what you mean. So I guessed the dominant term was n! and after dividing through by that and using basic null sequences I ended up with 5 so it converges to 5 but if it were 0 for example it would diverge?2012-04-02
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    No, the sequence would converge to 0 then. A sequence diverges if and only if the limit does not exist. Examples: the sequence $\bigl((-1)^n\bigr)$ diverges; the sequence $(n^2)$ diverges; the sequence $({2^n\over n!})$ converges to $0$. The terms of your sequence can be written as $5+{n^9\over n!}-{2^n\over n!}\over1+{2^n\over n!}+{n^{100}\over n!}$. As $n$ gets larger and larger, the non constant bits get closer and closer to 0. Then, the entire term gets closer and closer to 5. So the sequence converges to 5.2012-04-02
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    oh right yeah I see cheers!2012-04-02
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    @DavidMitra: Care to add to answer, rather than letting the content lie in the comments? Please see: http://meta.math.stackexchange.com/questions/1559/dealing-with-answers-in-comments2012-04-02

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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.