Find the limit of the sequence $frac{2^{n-1} + 3^n} {4^n - 1}$
$begingroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
$endgroup$
add a comment |
$begingroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
$endgroup$
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
add a comment |
$begingroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
$endgroup$
I have done right here...
$$lim_{n to infty } frac{2^{n-1} + 3^n}{4^n - 1}.$$
I know that I have to eliminate the lowest exponents in each part which is $2^{n-1}$ at the numerator and $1$ at the denominator but how to eliminate it?
sequences-and-series limits exponential-function
sequences-and-series limits exponential-function
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
edited Jan 8 at 8:47
user445909
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
asked Jan 8 at 8:43
Supakorn SrisawatSupakorn Srisawat
625
625
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
add a comment |
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
1
1
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46
add a comment |
1 Answer
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$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
add a comment |
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$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
add a comment |
$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
add a comment |
$begingroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
$endgroup$
Just divide numerator and denominator by $4^{n}$. Can you now see that the limit is $0$?
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
answered Jan 8 at 8:46
Kavi Rama MurthyKavi Rama Murthy
75.2k53270
75.2k53270
add a comment |
add a comment |
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$begingroup$
Divide by $4^n$
$endgroup$
– bigant146
Jan 8 at 8:46