`Lt_(x->a) [f(x)+g(x)` exists implies ....

If ("lim")_(x->a)[f(x)g(x)] exists, then both ("lim")_(x->a)f(x)a n d("lim")_(x->a)g(x) exist.

If ("lim")_(x->a)[f(x)g(x)] exists, then both ("lim")_(x->a)f(x)a n d("lim")_(x->a)g(x) exist.

If lim_(x->a)(f(x)/(g(x))) exists, then

If lim_(x->a)[f(x)g(x)] exists, then both lim_(xtoa)f(x) and lim_(x->a)g(x) exist.

verify the statement true or false.If lim_( x to a ) [f(x) g(x)] exists, then both lim_( x to a ) f(x) and lim_( x to a ) g (x) exist.

If lim_(xtoa) {(f(x))/(g(x))} exists, then

True or False : If underset(x rarr a)Lt f(x)g(x) exists then both underset(xrarra^+)Lt f(X) and underset(xrarra^-) Lt g(x) exist separately.

If lim_(x rarr a)[f(x)g(x)] exists,then both lim_(x rarr a)f(x) and lim_(x rarr a)g(x) exist.

If ("lim")_(xtoa)[f(x)g(x)] exists, then both ("lim")_(xtoa)f(x)a n d("lim")_(xtoa)g(x) exist.

If f(x)=sgn(x)" and "g(x)=x^(3) ,then prove that lim_(xto0) f(x).g(x) exists though lim_(xto0) f(x) does not exist.