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Calculate limxn(x+a1)(x+a2)...(x+an)x

This is a problem which I found in a math textbook some time ago. I tried solving it but I did not manage to solve it as quickly as I wanted and so I lost patience. Therefore I posted it in MathSE. I liked quite a lot one of the solutions I got, so I wrote it on a sheet of paper and kept it.

Here is the full story...

Problem:

Find limxn(x+a1)(x+a2)...(x+an)x,

where x is a real variable and n is a fixed natural number n1.

Solution:

Let
f(x)=n(x+a1)(x+a2)(x+an)

We now calculate:

limxf(x)x=limxn(1+a1x)(1+a2x)(1+anx)=n1=1

We now form the below difference and we rework it a bit:

f(x)x=f(x)nxnn1k=0f(x)n1kxk=f(x)nxnf(x)n1+f(x)n2x++f(x)xn2+xn1

We note here that the numerator P(x)=f(x)nxn is a polynomial of degree n1 and it has a leading coefficient of c=a1+a2++an. The denominator is composed of the sum of n terms of the form f(x)n1kxk (for k=0,1,,n1).

So we divide the numerator and the denominator by xn1 and we get:

limxf(x)x=limxf(x)nxnxn1f(x)n1xn1+f(x)n2xn2++f(x)x+1=limxP(x)xn1f(x)n1xn1+f(x)n2xn2++f(x)x+1=c1+1++1n times=a1+a2++ann

The cool thing here is that the limit value happens to be the arithmetic mean of the numbers a1,a2,an.


See also Math SE question/solution here




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