document.write( "Question 361546: Please help me solve this equation: Evaluate the following limits or explain why they do not exist.
\n" ); document.write( "a. \"+lim%28x-%3E0%2C+%28xcosx-sinx%29%2Fx+%29+\"
\n" ); document.write( "b. \"+lim%28x-%3Einfinity%2C+%28e%5Ex-2%5Ex%29%2Fx+%29+\"
\n" ); document.write( "c. \"+lim%28x-%3E1%2C++%28x%2F%28x-1%29+-+%281%2Flnx%29+%29+%29+\"
\n" ); document.write( "

Algebra.Com's Answer #257763 by robertb(5830)\"\" \"About 
You can put this solution on YOUR website!
a. Putting x = 0 produces the indeterminate form 0/0. But for this problem, there is NO need to use L'Hopital's rule as shown below:
\n" ); document.write( "But
\n" ); document.write( "= \"lim%28x-%3E0%2C+%28cosx%29%29-lim%28x-%3E0%2C+sinx%2Fx%29\" = 1 -1 = 0. \r
\n" ); document.write( "\n" ); document.write( "b. Putting x = 0 produces the indeterminate form 0/0.
\n" ); document.write( "Now, by direct application of L'Hopital's rule. This does NOT produce an indeterminate form, so the limit is \"e%5E0+-+%28ln2%29%2A2%5E0+=+1-ln2\".\r
\n" ); document.write( "\n" ); document.write( "c. \"+lim%28x-%3E1%2C++%28x%2F%28x-1%29+-+%281%2Flnx%29+%29+%29+\",
\n" ); document.write( " =\"+lim%28x-%3E1%2C+%28+%28xlnx-x%2B1%29%2F%28%28x-1%29%2Alnx%29%29+%29+\", combining fractions.
\n" ); document.write( "=\"+lim%28x-%3E1%2C+%28+lnx%2F%28lnx%2B%28x-1%29%2Fx%29%29+%29+\", applying L-H rule. (Gives 0/0)
\n" ); document.write( "=\"+lim%28x-%3E1%2C+%28+%28xlnx%29%2F%28xlnx%2Bx-1%29%29%29\", simplifying complex fractions, (Gives 0/0)
\n" ); document.write( "=\"+lim%28x-%3E1%2C+%28+%28lnx%2B1%29%2F%28lnx%2B2%29+%29+%29\", applying L-H rule (Not indeterminate anymore)
\n" ); document.write( "=\"1%2F2\"
\n" ); document.write( "
\n" );