Limit of a sequence

1003047608

Level: 
C
Choose the step to take first to efficiently evaluate the limit of the sequence \( \left( \frac{3n+2}{\sqrt{n^2-1}} \right)_{n=1}^{\infty} \).
We divide the numerator and the denominator by \( n \).
We take \( \sqrt n \) outside brackets in the numerator and the denominator.
We square the denominator.
We divide the numerator by \( n \).
We divide the denominator by \( n \).

1003047606

Level: 
C
The sequence \( \left( \sqrt n \left( \sqrt n-\sqrt{n-1} \right) \right)_{n=1}^{\infty} \) is:
convergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =\frac12 \)
convergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =0 \)
convergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =2 \)
divergent and \( \lim\limits_{n\to\infty} \sqrt n \left( \sqrt n-\sqrt{n-1} \right) =\infty \)
divergent and it does not have an infinite limit

1003047604

Level: 
C
Choose the correct computation of the limit. \[ L=\lim\limits_{n\to\infty} \left( \sqrt{n^2+3n}-2n \right) \]
\( L=\lim\limits_{n\to\infty}n\left( \sqrt{1+\frac3n}-2 \right) = -\infty \)
\( L= \infty-\infty=0 \)
\( L=\lim\limits_{n\to\infty}⁡(n-2n)=-\infty \)
\( L=\lim\limits_{n\to\infty} \left( n^2+3n-4n^2 \right) =-3 \)
\( L=\lim\limits_{n\to\infty}⁡\frac{n^2+3n-4n^2}{\sqrt{n^2+3n}+2n}=\infty \)

1003047602

Level: 
C
Choose the step to take first to efficiently evaluate the limit of the sequence \( \left(n-\sqrt{n^2-1} \right)_{n=1}^{\infty} \).
We expand with the expression \( n+\sqrt{n^2-1} \).
We expand with the expression \( n-\sqrt{n^2-1} \).
We expand with \( n \).
We multiply by the expression \( n+\sqrt{n^2-1} \).
We multiply by the expression \( n-\sqrt{n^2-1} \).
We substitute \( n=\infty \).