perplexus.info :: Calculus : Getting Cubic With Limits

Getting Cubic With Limits (Posted on 2008-02-09)

Determine all possible real t that satisfy this relationship:


Lim (5^p + 7^p + 11^p)^1/p = t³ - t + 5
p→∞

Submitted by K Sengupta

Rating: 4.0000 (1 votes)

Solution: (Hide)

The only possible value of t is 2.

EXPLANATION:

(5^p + 7^p + 11^p)^1/p
= 11*((5/11)^p + (7/11)^p + 1)^1/p)

Since each of (5/11)^p and (7/11)^p separately tends to zero whenever p tends to infinity, we must have:

Limit (5^p +7^p + 11^p)^1/p
p-> infinity

= 11*1 = 11

Hence, t^3 - t + 5 = 11
or, t^3 - t -6 = 0
or, (t-2)(t^2+t+3)=0

The only real root occurs at t=2, and so the required value of t is 2.

Comments: ( You must be logged in to post comments.)

Subject	Author	Date
Solution	Dej Mar	2008-02-09 21:42:01
Here it is	FrankM	2008-02-09 13:38:50

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