"Find the limits of the sequences: a)an=ln⁡nn+1 b)an=nsin⁡2n..." was answered by students like you

minikim38

Answered question

2022-01-24

Find the limits of the sequences:
a)

a_{n} = \frac{\ln n}{n + 1}

a_{n} = n \sin \frac{2}{n}

(n = 1, 2, 3, \dots)

Brynn Ortiz

Beginner2022-01-25Added 12 answers

(a)

a_{n} = \frac{\ln n}{n + 1}

We can write it as

lim_{n \to \infty} \frac{\ln n}{n + 1}

Solve the limit

lim_{n \to \infty} \frac{\ln (n)}{n + 1}

= \frac{\ln (\infty)}{(\infty) + 1} = \frac{\infty}{\infty} \dots .

(indeterminate form)
Apply the LHospital

Troy Sutton

Beginner2022-01-26Added 13 answers

a_{n} = n \sin \frac{2}{n}

Write as:

lim_{n \to \infty} \frac{\sin \frac{2}{n}}{\frac{1}{n}}

= \frac{\sin (\frac{2}{\infty})}{\frac{1}{\infty}} = \frac{0}{0}

(indeterminate form)
Now, apply the Hospital Rule

lim_{n \to \infty} \frac{\sin \frac{2}{n}}{\frac{1}{n}} = lim_{n \to \infty} \frac{\frac{d}{d n} (\sin \frac{2}{n})}{\frac{d}{d n} (\frac{1}{n})}

= lim_{n \to \infty} \frac{- \frac{2 \cos (\frac{2}{n})}{n^{2}}}{- \frac{1}{n^{2}}}

= lim_{n \to \infty} \frac{2 \cos (\frac{2}{n})}{n^{2}} \times \frac{n^{2}}{1}

lim_{n \to \infty} 2 \cos (\frac{2}{n})

= 2 \cos (\frac{2}{\infty})

= 2 \cos (0)

= 2

Hence, the answer is:

lim_{n \to \infty} n \sin \frac{2}{n} = 2

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