If `lim_(x rarr 0) (ax e^x-b ln(1+x)+c x e^-x)/(x^2 sinx)=2,` then `a+b+c=`

If lim_(x rarr0)(ax^(2)e^(x)-b log_(e)(1+x)+cxe^(-x))/(x^(2)sin x)=1 ,then 16(a^(2)+b^(2)+c^(2)) is equal to:

Find the constant a b and c such that lim_(x rarr0)(axe^(x)-b log(1+x)+cxe^(-x))/(x^(2)sin x)=2

lim_(x rarr0)(e^(x)-1)/(log(1+x))

lim_(x rarr 0) 2/x log (1+x) =

If lim_(xrarr0) frac{ax^2e^x-b log_e(1+x)+cxe^(-x)}{x^2 sinx}=1 then the value of 16(a^2+b^2+c^2) is

If lim_(x rarr0)(ae^(x)+b cos x+ce^(-x))/(e^(2x)-2e^(x)=1)=4, then a=2 b.b=-4 c.c=2 d.a+b+c=-8

lim_(x rarr 0) (e^x+e^-x-2 cosx)/(x sinx)

lim_(x rarr0)(e^(3x)-1)/(log(1+5x))

lim_(x rarr0)(e^(x)+log(1+x)-(1-x)^(-2))/(x^(2))