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

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

Evaluate underset(xrarr0)lim (e^(2x)-1)/(sinx) .

Evaluate : underset(xrarr0)"lim"((e^(x)-1)log(1+x))/(sinx)

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

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

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

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

underset(xrarr0)lim(xe^(x)-sinx)/(x) is equal to