Verify that $\displaystyle \lim_{n \rightarrow \infty} \frac{n + 3}{n^3 + 4} =0$ Hi every one we just started doing sequences in my calculus class, and we were given a couple practice problems that use the definition of a sequence, which is: for $\forall \epsilon > 0, \exists N > 0$ such that $ |Sn - l| < \epsilon$  $  \forall n>N.$  This is one of the practice problems that i need help with:  $\displaystyle \lim_{n \rightarrow \infty} \frac{n + 3}{n^3 + 4} =0$ We werent really given any examples to go off of so any help on this would be great! Thank You
 A: $\displaystyle\lim_{n\to\infty} \dfrac{n+3}{n^3+4}
=\lim_{n\to\infty} \dfrac{\dfrac1{n^2}+\dfrac{3}{n^3}}{1+\dfrac{4}{n^3}}$
Note that both the numerator and denominator have finite limits (with the denominator one being non-zero).
A: Note that $$0\leqslant \frac{n+3}{n^3+4}\leqslant \frac{n+3}{n^3}=\frac 1{n^2}+\frac 3{n^3}\leqslant \frac 1n+\frac 3 n=\frac  4n$$ Now squeeze.
A: Use the same as if it were $$\displaystyle\lim_{x\to\infty} \dfrac{x+3}{x^3+4}$$ The limit is given by the ratio of the highest order term in the numerator ($x$ here) to the highest order term in the denominator  ($x^3$ here). So, the asymptotic behavior is $\frac{1}{x^2}$ and $x$ goes to infinity.  
To make it clearer, factor $x$ in the numerator and $x^3$ in the denominator (these are the highest powers). So, the numerator write $x(1+3/x)$ and the denominator $x^3(1+4/x^3)$
; when $x$ goes to large values, terms such as $1/x$ and still more $4/x^3$ are unsignificant when compared to $1$; so you can forget them and, again, jus focus on what is left for large $x$'s, that is to say $x/x^3=1/x^2$
