Find the limits of the following sequences ({cos⁡((2n+1)π2)})n=1∞ ({πne2n+1})n=1∞ ({n2+3n3+n2−1})n=1∞ ({nsin⁡πn})n=1∞

1)({cos⁡((2n+1)π2)})n=1∞ limn→∞(cos⁡(2n+1)π2) =cos⁡(∞)π2 =cos⁡∞=O≠ (∵cos⁡∞=not unqiue) =limn→∞[cos⁡(2n+1)π2]=o≠ 2)({πne2n+1})n=1∞ limn→∞πne2n+1 limn→∞πn(e2)n×e limn→∞(πne2)n×1e =(πe2)∞×1e=0 =limn→∞πne2n+1=0

(c)({n2+3n3+n2−1})n=1∞ limn→∞nsin⁡πn sin⁡πn1n limn→∞(sin⁡πnπn) =π×1 =π (∵limx→∞sin⁡1x1x)=1 =limn→∞(nsin⁡πn)=π (d)({nsin⁡πn})n=1∞ limn→∞n2+3n3+n2−1 limn→∞n2(1+3n2)n3(1+1n−1n3) limn→∞1n(1+3n2)1+1n−1n3 1∞(1+3∞)1+1∞−1∞ =0×(1+0)1+0−0 =0

1)({cos⁡((2n+1)π2)})n=1∞ limn→∞(cos⁡(2n+1)π2) =cos⁡(∞)π2 =cos⁡∞=O≠ (∵cos⁡∞=not unqiue) =limn→∞[cos⁡(2n+1)π2]=o≠ 2)({πne2n+1})n=1∞ limn→∞πne2n+1 limn→∞πn(e2)n×e limn→∞(πne2)n×1e =(πe2)∞×1e=0 =limn→∞πne2n+1=0

(c)({n2+3n3+n2−1})n=1∞ limn→∞nsin⁡πn sin⁡πn1n limn→∞(sin⁡πnπn) =π×1 =π (∵limx→∞sin⁡1x1x)=1 =limn→∞(nsin⁡πn)=π (d)({nsin⁡πn})n=1∞ limn→∞n2+3n3+n2−1 limn→∞n2(1+3n2)n3(1+1n−1n3) limn→∞1n(1+3n2)1+1n−1n3 1∞(1+3∞)1+1∞−1∞ =0×(1+0)1+0−0 =0

"Find the limits of the following sequences ({cos⁡((2n+1)π2)})n=1∞..." solved and explained

Ethen Wong

Answered question

2022-01-24

Find the limits of the following sequences

{({\cos ((2 n + 1) \frac{π}{2})})}_{n = 1}^{\infty}

{({\frac{π^{n}}{e^{2 n + 1}}})}_{n = 1}^{\infty}

{({\frac{n^{2} + 3}{n^{3} + n^{2} - 1}})}_{n = 1}^{\infty}

{({n \sin \frac{π}{n}})}_{n = 1}^{\infty}

Answer & Explanation

waijazar1

Beginner2022-01-25Added 13 answers

1) ${({\cos ((2 n + 1) \frac{π}{2})})}_{n = 1}^{\infty}$
$lim_{n \to \infty} (\cos (2 n + 1) \frac{π}{2})$
$= \cos (\infty) \frac{π}{2}$
$= \cos \infty = O \neq$ $(∵ \cos \infty = n o t u n q i u e)$
$= lim_{n \to \infty} [\cos (2 n + 1) \frac{π}{2}] = o \neq$
2) ${({\frac{π^{n}}{e^{2 n + 1}}})}_{n = 1}^{\infty}$
$lim_{n \to \infty} \frac{π^{n}}{e^{2 n + 1}}$
$lim_{n \to \infty} \frac{π^{n}}{{(e^{2})}^{n} \times e}$
$lim_{n \to \infty} {(\frac{π^{n}}{e^{2}})}^{n} \times \frac{1}{e}$
$= {(\frac{π}{e^{2}})}^{\infty} \times \frac{1}{e} = 0$
$= lim_{n \to \infty} \frac{π^{n}}{e^{2 n + 1}} = 0$

bemolizisqt

Beginner2022-01-26Added 16 answers

(c)

{({\frac{n^{2} + 3}{n^{3} + n^{2} - 1}})}_{n = 1}^{\infty}

lim_{n \to \infty} n \sin \frac{π}{n}

\frac{\sin \frac{π}{n}}{\frac{1}{n}}

lim_{n \to \infty} (\frac{\sin \frac{π}{n}}{\frac{π}{n}})

= π \times 1

= π

(∵ lim_{x \to \infty} \frac{\sin \frac{1}{x}}{\frac{1}{x}}) = 1

= lim_{n \to \infty} (n \sin \frac{π}{n}) = π

(d)

{({n \sin \frac{π}{n}})}_{n = 1}^{\infty}

lim_{n \to \infty} \frac{n^{2} + 3}{n^{3} + n^{2} - 1}

lim_{n \to \infty} \frac{n^{2} (1 + \frac{3}{n^{2}})}{n^{3} (1 + \frac{1}{n} - \frac{1}{n^{3}})}

lim_{n \to \infty} \frac{\frac{1}{n} (1 + \frac{3}{n^{2}})}{1 + \frac{1}{n} - \frac{1}{n^{3}}}

\frac{\frac{1}{\infty} (1 + \frac{3}{\infty})}{1 + \frac{1}{\infty} - \frac{1}{\infty}}

= \frac{0 \times (1 + 0)}{1 + 0 - 0}

= 0

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