`lim_(x rarr 1) ln(1-lnx)/(lnx^2)=`

Find the following limits: lim_(xrarr1) (lnx)/(x^2-1)

Let a= lim _(x rarr 1) (x/(lnx)-1/(xln x)), b = lim _(x rarr 0) ((x^(3)-16x)/(4x+x^(2))), c= lim _(x rarr 0) ((ln(1+sinx))/x) & d = lim _(x rarr -1) ((x+1)^(3))/(3([sin (x+1) - (x+1)])) Then [[a,b],[c,d]] is

Let a= lim _(x rarr 1) (x/(lnx)-1/(xln x)), b = lim _(x rarr 0) ((x^(3)-16x)/(4x+x^(2))), c= lim _(x rarr 1) ((ln(1+sinx))/x) and c = lim _(x rarr -1) ((x+1)^(3))/([sin (x+1) - (x+1)]) then [[a,b],[c,d]] is

Let a= lim _(x rarr 1) (x/(lnx)-1/(xln x)), b = lim _(x rarr 0) ((x^(3)-16x)/(4x+x^(2))), c= lim _(x rarr 1) ((ln(1+sinx))/x) & d = lim _(x rarr -1) ((x+1)^(3))/([sin (x+1) - (x+1)]) Then [[a,b],[c,d]] is

Let a= lim _(x rarr 1) (x/(lnx)-1/(xln x)), b = lim _(x rarr 0) ((x^(3)-16x)/(4x+x^(2))), c= lim _(x rarr 1) ((ln(1+sinx))/x) & d = lim _(x rarr -1) ((x+1)^(3))/([sin (x+1) - (x+1)]) Then [[a,b],[c,d]] is

lim_(x rarr0)(log(1+x))/(x)=1

lim_(x rarr0)(ln(1+x)^(1+x))/(x^(2))-(1)/(x)

Find the following limits: lim_(xrarr1)(1/(x-1)-1/(lnx))

lim_(x rarr0)(log(1-(x)/(2)))/(x)

Using the L .Hospital rule find limits of the following functions : lim_(x to 1)(1/(ln x)-x/(lnx))