`underset(xtoa)lim(log(x-a))/(log(e^(x)-e^(a)))`

lim_(xtoa) (log(x-a))/(log(e^(x)-e^(a)))

underset (x rarr infty) lim(log x^(x))/(x)

Evaluate: underset(xrarr0)lim((log(1-x))/((x)))

underset(x rarr0)lim x^(2)log e^(x)

underset(x rarr oo)(lim) (x^(2))/(e^(x)) is

e ^(x log a)e^(x)

lim_(xto0)(log(1+3x^(2)))/(x(e^(5x)-1)) =

Evaluate underset(x rarr oo)(lim) (1 + x)^((1)/(log x))

If int((log_(ex)e)(log_(e^(2)x)e)log_(e^(3)x) e))1/x dx = Alog |1+logx|+B log |2+log x|+C log |3+ logx| +D then A-B+C is equal to

lim_(xtooo){x[log(x+a)-log(x)]}