Find the limit of the sequence given by $$\frac{10+12n+20n^4}{7n^4 + 5n^3 - 20}$$
I think the answer is $\frac{20}{7}$ after dividing, but is that right?
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Yes it is, because $$\frac{20n^4+12n+10}{7n^4 + 5n^3 - 20}=\frac{20+\frac{12}{n^3}+\frac{10}{n^4}}{7+\frac5n-\frac{20}{n^4}}\to\frac{20+0+0}{7+0-0}=\frac{20}7$$ by the arithmetic of limits.
You should see this method works in general for a fraction of two polynomials, just divide $n^d$, where $d$ is the degree of the denominator.
user2345215
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Yes right!
$$\frac{10+12n+20n^4}{7n^4+5n^3-20}=\frac{\frac{10}{n^4}+\frac{12}{n^3}+20}{7+\frac5n-\frac{20}{n^4}}$$ and pass to the limit.
Remark: There's a simple way to find the limit at $\infty$ for a fraction of two polynomials: it suffices to take the monomial of highest degree for the numerator and denominator, hence applying this we find
$$\lim_{n\to\infty}\frac{10+12n+20n^4}{7n^4+5n^3-20}=\lim_{n\to\infty}\frac{20n^4}{7n^4}=\frac{20}7$$
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Thanks! And one more question suppose there was a n in the denominator was under a square root, would that mean it diverges? – Amina Apr 19 '14 at 18:10
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We say that a sequence diverges if it hasn't a limit or tends to $\infty$. – Apr 19 '14 at 18:14
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Which limit do you want? As $n \to 0$ this goes to the limit $-\frac 12$.
I put this as an answer because the other answers were up, and there are alternatives to those.
Mark Bennet
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@user2345215 That doesn't stop $n$ decreasing to zero. I put this up, because a failure to specify is a mathematical error, and care needs to be taken with limits, even if these things do seem obvious. – Mark Bennet Apr 19 '14 at 18:11
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@user2345215 You are confusing $\mathbb{N}$ and $n$. $n$ is just a variable in this expression, and is almost certainly a real number (in this question). Not sure why this is downvoted. – Sidd Singal Apr 19 '14 at 18:14
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@mathguy Almost certainly not, see the word "sequence" in the title. This interpretation makes no sense. – user2345215 Apr 19 '14 at 18:17
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@user2345215 I missed the title too. But to be pedantic, the constant sequence where every term is zero tends to zero. I agree that is probably not what was contemplated. And sometimes we want $n\to - \infty$. – Mark Bennet Apr 19 '14 at 18:20
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Maybe I created this confusion, when I edited the post to be "clear". OP originally specified no text, just what was in the title, so I fixed it now. – user2345215 Apr 19 '14 at 18:21
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