document.write( "Question 438196: Let \"f%5Bn%5D%28x%29+=+nxe%5E%28-nx%5E2%29\", for x in [0,1]. Show whether or not
\n" ); document.write( "i) \"f%5Bn%5D%28x%29+\" converges for each x in [0,1].
\n" ); document.write( "ii) \"f%5Bn%5D%28x%29\" converges uniformly on [0,1].
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Algebra.Com's Answer #303110 by robertb(5830)\"\" \"About 
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i) Fix any x-value in the interval (0,1].
\n" ); document.write( "Now \"f%5Bn%5D%28x%29+=+nxe%5E%28-nx%5E2%29+=+%28nx%29%2Fe%5E%28nx%5E2%29\"\r
\n" ); document.write( "\n" ); document.write( "==> \r
\n" ); document.write( "\n" ); document.write( "=\"+x%2Alim%28n-%3Einfinity%2C+1%2F%28x%5E2e%5E%28nx%5E2%29%29%29\" by L'Hopital's rule.\r
\n" ); document.write( "\n" ); document.write( "=\"%281%2Fx%29%2Alim%28n-%3Einfinity%2C+1%2Fe%5E%28nx%5E2%29%29+=+%281%2Fx%29%2A0+=+0\"\r
\n" ); document.write( "\n" ); document.write( "If x = 0, \"f%5Bn%5D%280%29+=+%28n%2A0%29%2Fe%5E0+=+0\" for all n.\r
\n" ); document.write( "\n" ); document.write( "Hence for all x in [0,1], the sequence \"f%5Bn%5D%28x%29\" exhibits pointwise convergence, and \"lim%28n-%3Einfinity%2C+f%5Bn%5D%28x%29%29+=+0\".\r
\n" ); document.write( "\n" ); document.write( "ii) Now \"df%5Bn%5D%28x%29%2Fdx+=+%28n+-+2n%5E2x%5E2%29%2Fe%5E%28nx%5E2%29\".
\n" ); document.write( "Setting this derivative to 0 and solving for x, we get \"x%5E2+=+1%2F%282n%29\"
\n" ); document.write( "<==> \"x+=+1%2Fsqrt%282n%29\"
\n" ); document.write( "Implementing the 1st derivative test using the test points \"1%2Fsqrt%282%28n%2B1%29%29\" and \"1%2Fsqrt%282%28n-1%29%29\", we find that there is (absolute) maximum at \"x+=+1%2Fsqrt%282n%29\".
\n" ); document.write( " (Remember that \"1%2Fsqrt%282%28n%2B1%29%29+%3C+1%2Fsqrt%282n%29+%3C+1%2Fsqrt%282%28n-1%29%29\".)
\n" ); document.write( "Then \"f%5Bn%5D%281%2Fsqrt%282n%29%29+=+sqrt%28n%2F%282e%29%29\". Hence as \"n+-%3E+infinity\", the maximum of the function goes to infinity, and so the sequence of functions \"f%5Bn%5D%28x%29\" does not exhibit uniform convergence.
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