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`        The answer is nth root of a1a2a3....an please explain it.`
5 months ago

```							hello karishma this is an ez prob.let L be the reqd limit. thenlog L= lim x tends to zero log(f(x))/xwhere f(x)= (a1^x+a2^x+.....+an^x)/nnote that  lim x tends to zero f(x)= 1so that if we let g(x)= f(x) – 1, then  lim x tends to zero g(x)= 0now 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 bso log L= (1/n)*(loga1 + loga2 + ….. + logan)= (1/n)*(loga1a2.....an)take anti logL= (a1a2.....an)^(1/n)kindly approve :)
```
5 months ago
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