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The answer is nth root of a1a2a3....an please explain it.

The answer is nth root of a1a2a3....an please explain it.

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Grade:11

1 Answers

Aditya Gupta
2081 Points
3 years ago
hello karishma this is an ez prob.
let L be the reqd limit. then
log L= lim x tends to zero log(f(x))/x
where f(x)= (a1^x+a2^x+.....+an^x)/n
note that  lim x tends to zero f(x)= 1
so that if we let g(x)= f(x) – 1, then  lim x tends to zero g(x)= 0
now log L= lim x tends to zero g(x)/x*lim x tends to zero log(1+g(x))/g(x)
now lim x tends to zero log(1+g(x))/g(x) can be solved by putting y= g(x), then as x tends to zero y also tends to zero s this lim becomes lim y tends to zero log(1+y)/y which equals 1 (standard result).
now we have log L= lim x tends to zero g(x)/x
= lim x tends to zero [(a1^x+a2^x+.....+an^x)/n – 1]/x
= lim x tends to zero [{(a1^x – 1)+(a2^x – 1)+.....+(an^x – 1)}/n]/x
= (1/n)*lim x tends to zero [(a1^x – 1)+(a2^x – 1)+.....+(an^x – 1)]/x
= (1/n)*lim x tends to zero (a1^x – 1)/x+(a2^x – 1)/x+.....+(an^x – 1)/x
= (1/n)* [ lim(a1^x – 1)/x + lim(a2^x – 1)/x + ….. + lim(an^x – 1)/x ]
now we use the standard result lim y tends to zero (b^y – 1)/y= log b
so log L= (1/n)*(loga1 + loga2 + ….. + logan)
= (1/n)*(loga1a2.....an)
take anti log
L= (a1a2.....an)^(1/n)
kindly approve :)

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