यदि फलन f(x) जहाँ `f(x)={{:((log(1+ax)-log(1-bx))/(x)",",x ne 0),(" "k",",x = 0):}x = 0` पर संतत है, तब k का मान ज्ञात कीजिए ।

Explanation

(i) `f(0) = k`
(ii) R.H.L. `=underset(x rarr 0^(+))(lim)f(x) = underset(h rarr 0)(lim)f(0+h)`
`=underset(h rarr 0)(lim)(log(1+a(0+h))-log(1-b(0+h)))/((0+h))`
`underset(h rarr 0)(lim)(log(1+ah)-log(1-bh))/(h)`
`=underset(h rarr 0)(lim)(log(1+ah))/(h)-underset(h rarr 0)(lim)(log(1-bh))/(h)`
`=a xx underset(h rarr 0)(lim)(log (1+ah))/(ah)-(-b)underset(h rarr 0)(lim)(log (1-bh))/((-b)h)`
`= a xx 1 - (-b) xx 1," "[because underset(x rarr 0)(lim)(log (1+x))/(x)=1]`
`rArr" ""R.H.L." = a + b`
(iii) L.H.L. `underset(x rarr 0^(-))(lim)f(x) = underset(h rarr 0)(lim)f(0-h)`
`underset(h rarr 0)(lim)(log(1+a(0-h))-log(1-b(0-h)))/((0-h))`
`=underset(h rarr 0)(lim)(log(1-ah)-log(1+bh))/(-h)`
`=underset(h rarr 0)(lim)(log(1-ah))/(-h)+underset(h rarr 0)(lim)(log(1+bh))/(h)`
`= a xx underset(h rarr 0)(lim)(log(1-ah))/((-a)h)+b xx underset(h rarr 0)(lim)(log(1+x))/(bh)`
`rArr" ""L.H.L." = a xx 1 + b xx 1 = a + b, " "[because underset(x rarr 0)(lim)(log(1+x))/(x)=1]`
चूँकि फलन f(x) बिन्दु x = 0 पर संतत है ।
`therefore" ""R.H.L.=L.H.L." = f(0)`
`rArr" "a + b =k`.