Question 438196
i) Fix any x-value in the interval (0,1].
Now {{{f[n](x) = nxe^(-nx^2) = (nx)/e^(nx^2)}}}

==> {{{lim(n->infinity, (nx)/e^(nx^2) ) = x*lim(n->infinity, n/e^(nx^2) ) }}}

={{{ x*lim(n->infinity, 1/(x^2e^(nx^2)))}}} by L'Hopital's rule.

={{{(1/x)*lim(n->infinity, 1/e^(nx^2)) = (1/x)*0 = 0}}}

If x = 0, {{{f[n](0) = (n*0)/e^0 = 0}}} for all n.

Hence for all x in [0,1], the sequence {{{f[n](x)}}} exhibits pointwise convergence, and {{{lim(n->infinity, f[n](x)) = 0}}}.

ii) Now {{{df[n](x)/dx = (n - 2n^2x^2)/e^(nx^2)}}}.
Setting this derivative to 0 and solving for x, we get {{{x^2 = 1/(2n)}}}
<==> {{{x = 1/sqrt(2n)}}}
Implementing the 1st derivative test using the test points {{{1/sqrt(2(n+1))}}} and  {{{1/sqrt(2(n-1))}}}, we find that there is (absolute) maximum at {{{x = 1/sqrt(2n)}}}. 
 (Remember that {{{1/sqrt(2(n+1)) < 1/sqrt(2n) < 1/sqrt(2(n-1))}}}.)
Then {{{f[n](1/sqrt(2n)) = sqrt(n/(2e))}}}. Hence as {{{n -> infinity}}}, the maximum of the function goes to infinity, and so the sequence of functions {{{f[n](x)}}} does not exhibit uniform convergence.