`lim_(xto1)(log x)/(x-1)` का मान कीजिये।

Explanation

माना `x=1+h` यदि `x=1,` तब `h=0`
(`because` जब फलन कि चर राशि बदलेगी तब रफलां कि सिमा भी बदलेगी)
अतः `xto1implieshto0`
अतः `underset(xto1)lim(logx)/(x-1)impliesunderset(hto0)lim(log (1+h))/(1+h-1)`
`=underset(h to 0)lim(log(1+h))/(h)`
हम जानते है कि
`log (1+x)=x-1/2x^(2)+1/3x^(3)-1/4x^(4)+...oo`
अतः `underset(hto0)lim (log (1+h))/(h)=underset(hto0)lim (h-1/2h^(2)+1/3h^(3)+1/3h^(3)-1/4h^(4)+...oo)/(h)`
`=underset(hto0)lim(h[1-1/2h+1/3h^(2)-1/4h^(3)+...oo])/(h)`
`=underset(hto0)lim [1-1/2h+1/3h^(2)-1/4h^(3)+...oo]`
`=1-0+0+...` (सिमा लेने पर)
`therefore underset(xto1)lim(log x)/(x-1)=1`