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If you had the following limit, $$\lim _{n\to \infty} \frac {4n^3}{3n^2+1}$$ How would you determine this?

I know that one would just take the coefficients of the highest power in the denominator and numerator so it would be $4/3$. But how would one show this mathematically?

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    It should be $\infty$, not $4/3$ – 2017-01-25
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    The "ratio of highest degree coefficients" tricl only works when the numerator and the denominator has _the same degree_. – 2017-01-25
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    oh this is what I was wondering about – 2017-01-25

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The usual technique is to divide top and bottom by the highest power of $n$ in the bottom, in this case $n^2$: $$ \frac{4n}{3 + \frac{1}{n^2}} = \frac{4}{3 + \frac{1}{n^2}}\times n $$

Now, as $n \to \infty$, $\frac{1}{n^2} \to 0$, so the bottom goes to $3$, while the top is $4n$, which goes to infinity, so the limit is infinity.

However, if the denominator were $3n^3 + 1$, then I want you to see that the answer is $\frac 43$.

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    that makes perfect sense, I was unaware that you had to divide by the highest in the bottom. I figured you would divide by the highest overall. Thanks a lot – 2017-01-25
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    @bjp409 You are welcome. – 2017-01-25
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    @bjp409 You can divide by the highest overall if you like. In this case, it would give you $\frac{4}{\frac3n +\frac{1}{n^3}}$. If we look at what happens when $n\to\infty$, we see that the numerator stays at $4$, but the denominator becomes smaller and smaller. We still conclude that the limit is $\infty$. Exactly which you should divide by is part personal preference and part what works best in each problem. – 2017-01-26
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Perhaps it's clearer if we look at $$ \lim_{n\to\infty}\frac{4n^3+3n+1}{3n^2+1} $$ The trick is to collect the highest power of $n$ in the numerator and the highest power of $n$ in the denominator: $$ \lim_{n\to\infty} \frac{n^3\bigl(4+\frac{3}{n^2}+\frac{1}{n^3}\bigr)} {n^2\bigl(3+\frac{1}{n^2}\bigr)}= \lim_{n\to\infty} n\frac{4+\frac{3}{n^2}+\frac{1}{n^3}} {3+\frac{1}{n^2}} $$ Now the fraction has limit $4/3$, but there's the factor $n$ so the limit is $\infty$.

Your case is the same, there's just one term in the numerator, which of course is the highest degree: $$ \lim_{n\to\infty}\frac{4n^3+3n+1}{3n^2+1}= \lim_{n\to\infty} \frac{n^3\cdot4} {n^2\bigl(3+\frac{1}{n^2}\bigr)}= \lim_{n\to\infty} n\frac{4}{3+\frac{1}{n^2}}=\infty $$

This also works for other cases: $$ \lim_{n\to\infty}\frac{2n^2+n-1}{n^3-n^2-n+1}= \lim_{n\to\infty} \frac{n^2\bigl(2+\frac{1}{n}-\frac{1}{n^2}\bigr)} {n^3\bigl(1-\frac{1}{n}-\frac{1}{n^2}+\frac{1}{n^3}\bigr)}= \lim_{n\to\infty} \frac{1}{n} \frac{2+\frac{1}{n}-\frac{1}{n^2}} {1-\frac{1}{n}-\frac{1}{n^2}+\frac{1}{n^3}} $$ Here the fraction has limit $2$, but the factor $1/n$ makes the limit to be $0$.

If the numerator and denominator have the same degree, it's similar: you just get a fraction which you can easily compute the limit of.

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Actually, the limit is $\infty$. It works like this: $\lim_{x\to+\infty}\frac{4n^3}{3n^2+1} = \lim_{x\to+\infty}\frac{n^3*(4)}{n^2*(3+\frac{1}{n^2})} = \lim_{x\to+\infty}\frac{n*(4)}{3+\frac{1}{n^2}}$. Because $\frac{1}{n^2} = 0$ (a number divided by an infinitely large number is equal to zero), we get $\lim_{x\to+\infty}\frac{n*4}{3}$. Now it should be clear that this results in $+\infty$.

Note: If it were $4n^2$ instead of $4n^3$, the limit would be $\frac{4}{3}$.