`underset(xrarr0)"lim"(log(1+4x))/(x)` =

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

Evaluate : underset(xrarr0)"lim"(1)/(x) .

underset(xrarr0)"lim"(e^(3x)-1)/(x) =

underset(xrarr0)"lim"(a^(2x)-1)/(x) =

Prove : underset(xrarr0)"lim""xcos"(1)/(x)=0

Find the limit (iF exists ) : underset(xrarr0)"lim"(|x|)/(x)

Using underset(xrarr0)"lim"(e^(x)-1)/(x)=1" deduce that ," underset(xrarr0)"lim"(a^(x)-1)/(x)=log_(e)a[agt0].

Evaluate underset(xrarr0)(lim)(sinx+log(1-x))/(x^2)

Evaluate : underset(xrarr0)"lim"(sin^(-1)x)/(2x)

underset(xrarr0)lim (cos x) /(x^(2))