9000063608
Level:
B
Find the following limit.
\[
\lim _{n\to \infty }\frac{2^{n} + 3^{n}}
{3^{n}}
\]
\(1\)
\(2\)
\(3\)
\(\infty \)
Limit of a sequence
1003047501
Level:
C
Find the following limit. \[ \lim\limits_{n\rightarrow\infty}(2\sqrt{n}-3) \]
\( \infty \)
\( 2 \)
\( -1 \)
\( 0 \)
\( -3 \)
1003047502
Level:
C
Find the following limit. \[ \lim\limits_{n\rightarrow\infty}\frac 4{2\sqrt{n}-3} \]
\( 0 \)
\( 2 \)
\( \frac12 \)
\( 4 \)
\( \infty \)
1003047505
Level:
C
Find the following limit. \[ \lim\limits_{n\rightarrow\infty}\Bigl(\frac5{\sqrt{n}}-\frac2{3^n} +\frac{(-1)^n}{2n^2-1}-7\Bigr) \]
\( -7 \)
\( 0 \)
\( -4 \)
\( \infty \)
\( -\infty \)
1003047601
Level:
C
Find the limit \( \lim\limits_{n\to\infty}\left(n-\sqrt{n-1}\right) \).
\( \infty \)
\( 0 \)
\( -\infty \)
\( 1 \)
\( \frac12 \)
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 \( n=\infty \).
1003047603
Level:
C
Find the limit \( \lim\limits_{n\to\infty}\left( \sqrt{4n^2+3n}-2n \right) \).
\( \frac34 \)
\( \infty \)
\( 0 \)
\( -\infty \)
\( \sqrt2 \)
1003047604
Level:
C
Choose the correct computation of the limit. \[ L=\lim\limits_{n\to\infty} \left( \sqrt{n^2+3n}-2n \right) \]
\( L=\lim\limits_{n\to\infty}n\left( \sqrt{1+\frac3n}-2 \right) = -\infty \)
\( L= \infty-\infty=0 \)
\( L=\lim\limits_{n\to\infty}(n-2n)=-\infty \)
\( L=\lim\limits_{n\to\infty} \left( n^2+3n-4n^2 \right) =-3 \)
\( L=\lim\limits_{n\to\infty}\frac{n^2+3n-4n^2}{\sqrt{n^2+3n}+2n}=\infty \)
1003047605
Level:
C
Find the limit \( \lim\limits_{n\to\infty} \left( \sqrt n-\sqrt{n-1} \right) \).
\( 0 \)
\( \infty \)
\( -\infty \)
\( -1 \)
\( \sqrt2 \)
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