Compute the limit of the sequences: (a)(1-\frac{3}{n})^{2n} (b)3^{1/n}

William Collins 2022-01-04 Answered
Compute the limit of the sequences:
\(\displaystyle{\left({a}\right)}{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}^{{{2}{n}}}\)
\(\displaystyle{\left({b}\right)}{3}^{{\frac{{1}}{{n}}}}\)

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Shawn Kim
Answered 2022-01-05 Author has 4997 answers
(a)
\(\displaystyle{\left({a}\right)}{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}^{{{2}{n}}}\)
\(\displaystyle\lim_{{{n}\rightarrow{d}}}{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}^{{{2}{n}}}\)
\(\displaystyle=\lim_{{{n}\rightarrow{d}}}{e}^{{{\ln{{\left[{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}^{{{2}{n}}}\right]}}}}}\)
\(\displaystyle=\lim_{{{n}\rightarrow{d}}}{e}^{{{2}{n}\times{\ln{{\left({1}-{\frac{{{3}}}{{{n}}}}\right\rbrace}}}}}\)
\(\displaystyle={e}^{{\lim_{{{n}\rightarrow{d}}}{2}{n}\times{\ln{{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}}}}}\)
\(\displaystyle={e}^{{{2}\lim_{{{n}\rightarrow{d}}}{n}\times{\ln{{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}}}}}\)
\(\displaystyle={e}^{{\lim_{{{n}\rightarrow{d}}}}}{\frac{{{\ln{{\left({1}-{\frac{{{3}}}{{{n}}}}\right)}}}}}{{\frac{{1}}{{n}}}}}\)
\(\displaystyle{e}^{{\lim_{{{n}\rightarrow{d}}}}}{\left({\frac{{{\frac{{-{1}}}{{{1}-{\frac{{{3}}}{{{n}}}}}}}{x}}}{{\frac{{1}}{{n}^{{{2}}}}}}}\right)}\)
\(\displaystyle{e}^{{2}}{x}{\left\lbrace\lim_{{{n}\rightarrow{d}}}\right\rbrace}{\frac{{-{3}}}{{{1}-\frac{{3}}{{n}}}}}\)
\(\displaystyle{e}^{{{2}{x}{\left(-{3}\right)}\lim_{{{n}\rightarrow{d}}}}}{\frac{{{1}}}{{{1}-\frac{{3}}{{n}}}}}\)
\(\displaystyle{e}^{{-{6}{x}}}{\frac{{{1}}}{{{1}-\frac{{3}}{{n}}}}}={e}^{{-{6}{x}{\frac{{{1}}}{{{1}-{0}}}}}}\)
\(\displaystyle{e}^{{-{6}\times{1}}}={e}^{{-{6}}}\)
(b)
\(\displaystyle{3}^{{\frac{{1}}{{n}}}}\)
\(\displaystyle\lim_{{{n}\rightarrow{d}}}{3}^{{\frac{{1}}{{n}}}}\)
\(\displaystyle{3}^{{{0}}}={1}\)
Not exactly what you’re looking for?
Ask My Question
0
 

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Relevant Questions

asked 2021-08-15

Find the limits of the following sequences.
1) \(\cos\bigg((2n+1)\frac{\pi}{2}\bigg)^{\infty}_{n=1}\)
2) \(\frac{\pi^{n}}{e^{2n+1}}^{\infty}\)

asked 2022-01-04
Find the limit of the sequences:
\(\displaystyle{a}_{{n}}={\frac{{{8}{n}^{{2}}+{1}{n}+{1}}}{{{7}{n}^{{2}}+{9}{n}+{9}}}}\)
asked 2021-05-28
Verify, using the definition of convergence of a sequence, that the following sequences converge to the proposed limit.
a) \(\lim \frac{2n+1}{5n+4}=\frac{2}{5}\)
b) \(\lim \frac{2n^3}{n^3+3}=0\)
c) \(\lim \frac{\sin (n^2)}{\sqrt[3]{n}}\)
asked 2022-01-06
Determine the limits of the following sequences:
a. \(\displaystyle{a}_{{n}}=\frac{{{3}{n}^{{{3}}}}}{{{n}^{{{3}}}+{1}}}\)
b. \(\displaystyle{b}_{{n}}={\left(\frac{{{n}+{5}}}{{{n}}}\right)}^{{{n}}}\)
c. \(\displaystyle{c}_{{n}}={n}^{{\frac{{1}}{{n}}}}\)
asked 2021-05-28
Substitute n=1, 2, 3, 4, 5 and write the first five terms of the sequence
\(\left\{\frac{(-1)^{n-1}x^{2n-1}}{(2n-1)!}\right\}\)
asked 2021-08-19
Determine the convergence or divergence of the following sequences. If it converges give it its limit
\(\displaystyle{a}_{{n}}={\frac{{{\ln{{\left({n}^{{3}}\right)}}}}}{{{2}{n}}}}\)
asked 2021-06-10
Substitute n=1, 2, 3, 4, 5 and find the first five sequences in sequence
\(\left\{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\dotsm+\frac{1}{2^{n}}\right\}\)
...