Letf be a continuous function on R such that f(1π)=(sinen)e−n2+n2n2+1.  Then the value of f(0)  is

As f  is continuous so f(0)=limx→0f(x) f(0)=limn→∞f(1/4n)=limn→∞((sinen)e−n2+11+1/n2) =0+1=1