Limit of a sequence

1003047301

Level: 
A
Which of the following expressions describes the correct calculation of the limit? \[ L=\lim\limits_{n\to\infty}\frac{n^3+2n-3}{2n^3+5}\]
\( L=\lim\limits_{n\to\infty}\frac{1+\frac2{n^2}-\frac3{n^3}}{2+\frac{5}{n^3}}= \frac12 \)
\( L=\frac{\infty^3+2\cdot\infty-3}{2\cdot\infty^3+5}=\infty \)
\( L=\frac{\infty^3+2\cdot\infty-3}{2\cdot\infty^3+5}=0 \)
\( L=\lim\limits_{n\to\infty}\frac{n\left(n^2+2\right)-3}{2n^3+5}= -\frac35 \)
\( L=\lim\limits_{n\to\infty}\frac{\left(n^2+3\right)(n-1)}{2\left(n^3+\frac52\right)}= 0 \)

1003047302

Level: 
A
Choose the best first step to take in order to evaluate the limit of the following sequence. \[ \left( \frac{4n^5+n^4-n^3+2}{7n^4-2n^2+7n} \right)^{\infty}_{n=1} \]
We factor out \( n^4 \) in the numerator and in the denominator separately.
We factor out \( n^5 \) in the numerator and in the denominator separately.
We factorize a polynomial in the denominator.
We divide the denominator by \( n^4 \).
We divide the numerator by \( n^5 \).

1003047305

Level: 
A
The sequence \[ \left(\frac{12n^3+5n+1}{2n^3-6}\right)_{n=1}^{\infty} \]
is convergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=6 \).
is convergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=0 \).
is convergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=12 \).
is divergent and \( \lim\limits_{n\to\infty}\frac{12n^3+5n+1}{2n^3-6}=\infty \).
has no limit.