document.write( "Question 1118241: Use a graph of f or some other method to determine what, if any, value to assign to f(a) to make f continuous at x = a. HINT [See Example 2.] (If there is no such value, enter NONE.)
\n" ); document.write( "f(x) = 2-2e^x/x; a=0\r
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Algebra.Com's Answer #733521 by greenestamps(13200)\"\" \"About 
You can put this solution on YOUR website!


\n" ); document.write( "A couple of preliminary comments....

\n" ); document.write( "(1) Do you really think it is any help to us to include \"HINT [See Example 2.]\" in your message to us?!

\n" ); document.write( "(2) The problem as you show it is of little interest, because it is impossible to define f(a) to make f continuous at x=0:

\n" ); document.write( "\"f%280%29+=+2-2e%5E0%2F0+=+2-1%2F0\" which is undefined.

\n" ); document.write( "If you are working on a problem like this, then your knowledge of mathematics should be sufficient that you would know appropriate use of parentheses is critical.

\n" ); document.write( "The function you are asking about is, in fact, \"f%28x%29+=+%282-2e%5Ex%29%2Fx\"; not \"f%28x%29+=+2-2e%5Ex%2Fx\".

\n" ); document.write( "At x=0, the function value is the indeterminate form 0/0.

\n" ); document.write( "So use l'Hospital's rule.

\n" ); document.write( "limit as x approaches 0 of \"%282-2e%5Ex%29%2Fx\"

\n" ); document.write( "is equal to

\n" ); document.write( "limit as x approaches 0 of \"-2e%5Ex%2F1\" = -2e^0 = -2

\n" ); document.write( "A graphing calculator will confirm this answer.

\n" ); document.write( "\"graph%28400%2C400%2C-5%2C5%2C-10%2C2%2C%282-2e%5Ex%29%2Fx%29\"
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