1003047410
Level:
B
Find the limit \(\lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{3\cdot5^n-1} \).
\( 0 \)
\( \infty \)
\( \frac23 \)
\( -5 \)
\( 4 \)
Limit of a sequence
1003047409
Level:
B
The sequence \( \left(\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}\right)_{n=1}^{\infty} \) is:
divergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=\infty \)
convergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=\frac12 \)
convergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=\frac14 \)
convergent and \( \lim\limits_{n\rightarrow\infty}\frac{2\cdot3^n+4^n+5}{4\cdot3^n-1}=0 \)
divergent and it does not have an infinite limit
1003047408
Level:
B
Choose the best first step to simplify and calculate the limit \(\lim\limits_{n\rightarrow\infty}\frac{3^n+4^{n-1}}{3^n+4^{n+1}} \).
We divide the numerator and the denominator by \( 4^n \).
We divide the numerator and denominator by \( 3^n \).
We substitute \(n=\infty \).
We take \( 3^n \) outside the brackets in the numerator and the denominator.
We take \( 4 \) outside the brackets in the numerator and the denominator.
1003047407
Level:
B
Find the limit \( \lim\limits_{n\rightarrow\infty}\frac{3^{n+1}+4^{n+1}}{3^{n-1}+4^{n-1}}\).
\( 16 \)
\( \frac1{16} \)
\( 0 \)
\( \infty \)
\( 1 \)
1003047406
Level:
B
Choose the correct formula to calculate the limit. \[ L=\lim\limits_{n\rightarrow\infty}\frac{3^{n+1}+4^n}{2^n} \]
\( L=\lim\limits_{n\rightarrow\infty}\left(3\cdot\left(\frac32\right)^n+2^n\right)=\infty \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{2^n\left(3\cdot\left(\frac32\right)^n+2^n \right)}{2^n}=0 \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{3^n \left(3+\left(\frac43\right)^n\right)}{2^n}=0 \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{7^{n+1}}{2^n}=\infty \)
\( L=\frac{3^{\infty+1}+4^{\infty}}{2^{\infty}} =\frac72 \)
1003047405
Level:
B
The sequence \( \left(\frac{3^n-4^{n-1}}{4^n}\right)_{n=1}^{\infty} \) is:
convergent and \(\lim\limits_{n\rightarrow\infty}\frac{3^n-4^{n-1}}{4^n}=-\frac14 \)
convergent and \(\lim\limits_{n\rightarrow\infty}\frac{3^n-4^{n-1}}{4^n}=\frac14 \)
convergent and \(\lim\limits_{n\rightarrow\infty}\frac{3^n-4^{n-1}}{4^n}=-1 \)
convergent and \(\lim\limits_{n\rightarrow\infty}\frac{3^n-4^{n-1}}{4^n}=0 \)
divergent
1003047404
Level:
B
Find the limit \( \lim\limits_{n\rightarrow\infty}\frac{5^{n+1}+6^n}{5^n+6^{n+1} } \).
\( \frac16 \)
\( \frac56 \)
\( 1 \)
\( 0 \)
\( \infty \)
1003047403
Level:
B
Find the limit \( \lim\limits_{n\rightarrow\infty}\frac{5^{n+1}+6^{n+1}}{5^{n+1}+6^n} \).
\( 6 \)
\( 5 \)
\( \frac56 \)
\( 0 \)
\( \infty \)
1003047402
Level:
B
Choose the best first step to simplify and calculate the limit of the following sequence. \[ \left(\frac{3\cdot5^n+2\cdot6^n}{2\cdot5^n+4\cdot6^n}\right)_{n=1}^{\infty} \]
We take \( 6^n \) outside the brackets in the numerator and the denominator.
We take \( 5^n \) outside the brackets in the numerator and denominator.
We divide the numerator and denominator by \( 5^n \).
We divide the numerator by \( 6^n \).
We divide the denominator by \( 6^n \).
1003047401
Level:
B
Choose the correct formula to calculate the limit. \[ L=\lim\limits_{n\rightarrow\infty}\frac{3\cdot5^n+2\cdot6^n}{2\cdot5^n+4\cdot7^n } \]
\( L=\lim\limits_{n\rightarrow\infty} \frac{3\cdot\left(\frac57\right)^n+2\cdot\left(\frac67\right)^n}{2\cdot\left(\frac57\right)^n+4} =0 \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{3\cdot\left(\frac56\right)^n+2}{2\cdot\left(\frac57\right)^n+4}=\frac12 \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{3+2\cdot\left(\frac65\right)^n}{2+4\cdot\left(\frac75\right)^n } =\frac32 \)
\( L=\frac{3\cdot5^{\infty}+2\cdot6^{\infty}}{2\cdot5^{\infty}+4\cdot7^{\infty}}=\infty \)
\( L=\lim\limits_{n\rightarrow\infty}\frac{3\cdot\left(\frac57\right)^n+2\cdot\left(\frac67\right)^n}{2\cdot\left(\frac57\right)^n+4\cdot\left(\frac77\right)^n}=\frac56 \)