find limit of a sequenceI have a sequence:Xn=cos⁡((3n+πn3n+(π−14)n)1n)I have to find the limit when n→∞ where n∈N.Which is the best way to find the answer ?

3n+πn3n+(π−{1over4})n=(π3)n1+(3π)n1+(π−143)nNow π−14&lt;3&lt;π, so 1+(3π)n→1 and 1+(π−143)n→1. Thus(3n+πn3n+(π−14)n)1n→π3Taking the cosine gives the limit cos⁡(π3)=12

"find limit of a sequenceI have a sequence:Xn=cos⁡((3n+πn3n+(π−14)n)1n)I..." solved and explained

High School
Calculus and Analysis
Calculus 1
Limits and continuity

Ormezzani6cuu

Answered question

2022-04-12

find limit of a sequence
I have a sequence:

X_{n} = \cos ({(\frac{3^{n} + π^{n}}{3^{n} + {(π - \frac{1}{4})}^{n}})}^{\frac{1}{n}})

I have to find the limit when

n \to \infty

where

n \in N

.
Which is the best way to find the answer ?

Answer & Explanation

izvozna39g0

Beginner2022-04-13Added 9 answers

\frac{3^{n} + π^{n}}{3^{n} + {(π - {1 o v e r 4})}^{n}} = {(\frac{π}{3})}^{n} \frac{1 + {(\frac{3}{π})}^{n}}{1 + {(\frac{π - \frac{1}{4}}{3})}^{n}}

Now

π - \frac{1}{4} < 3 < π

, so

1 + {(\frac{3}{π})}^{n} \to 1

and

1 + {(\frac{π - \frac{1}{4}}{3})}^{n} \to 1

. Thus

{(\frac{3^{n} + π^{n}}{3^{n} + {(π - \frac{1}{4})}^{n}})}^{\frac{1}{n}} \to \frac{π}{3}

Taking the cosine gives the limit

\cos (\frac{π}{3}) = \frac{1}{2}

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