How to find the limit of the sequence with n-th termxn=(2n+3n2)n

As usual when the variable appears in the exponent, write it using the rule ab=exp⁡(bln⁡a)exp⁡(nln⁡(2n+3n2))andnln⁡(2n+3n2=nln⁡(2n+3)−2nln⁡(n)=nln⁡(n)(ln⁡(2+3n)−2)It is now easy to conclude: nln⁡(n)→∞ and ln⁡(2+3n)−2→ln⁡2−2&lt;0, so nln⁡(2n+3n2)→−∞ and exp⁡(nln⁡(2n+3n2))→0To have a chance to apply l’Hospital rule, try to put x=1n and study the limit in 0, but I think this is not very useful here, as the above method is rather straightforward.

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High School
Calculus and Analysis
Calculus 1
Limits and continuity

Sullivan Pearson

Answered question

2022-04-22

How to find the limit of the sequence with n-th term

x_{n} = {(\frac{2 n + 3}{n^{2}})}^{n}

Answer & Explanation

2sze1c1se3nh

Beginner2022-04-23Added 17 answers

As usual when the variable appears in the exponent, write it using the rule

a^{b} = \exp (b \ln a)

\exp (n \ln (\frac{2 n +}{3 n^{2}}))

and

n \ln (\frac{2 n + 3}{n^{2}} = n \ln (2 n + 3) - 2 n \ln (n)

= n \ln (n) (\ln (2 + \frac{3}{n}) - 2)

It is now easy to conclude:

n \ln (n) \to \infty

and

\ln (2 + \frac{3}{n}) - 2 \to \ln 2 - 2 < 0

, so

n \ln (\frac{2 n + 3}{n^{2}}) \to - \infty

and

\exp (n \ln (\frac{2 n + 3}{n^{2}})) \to 0

To have a chance to apply l’Hospital rule, try to put

x = \frac{1}{n}

and study the limit in 0, but I think this is not very useful here, as the above method is rather straightforward.

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