Limit of a sequence

1003047309

Level: 
A
The sequence \[ \left(\frac{3n^5+2n^3+1}{n^3+3}\right)_{n=1}^{\infty} \]
is divergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=\infty \).
is convergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=0 \).
is convergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=3 \).
is divergent and \( \lim\limits_{n\to\infty}\frac{3n^5+2n^3+1}{n^3+3}=-\infty \).
has no limit.

1003047308

Level: 
A
Choose the best first step to take in order to evaluate the limit of the following sequence. \[ \left(\frac{3n^2-2n+4}{8n^2+13n+2}\right)_{n=1}^{\infty} \]
We divide the numerator and the denominator by \( n^2 \).
We divide the numerator and the denominator by \( n \).
We substitute \( n=\infty \).
We factor out \( n \) in the numerator and in the denominator separately.
We factor out \( 8 \) in the numerator and in the denominator separately.