document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=437663>Question 437663</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Give the equation of the horizontal asymptote, if any, of the function.\r<BR>\n" );
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document.write( "G(x) = x(x-1)/x^3+49x\r<BR>\n" );
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document.write( "A) y = 0     B) y = 1     C) x = 0, x = -49     D) None\r<BR>\n" );
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document.write( "I worked it out like this: x(x-1)/x^3+49x = x^2-x/x(x^2+49)     x^2/x^2  y = 1<BR>\n" );
document.write( "However, I am not sure if that is correct. What are the correct steps to solve this and what is the right answer?\r<BR>\n" );
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document.write( "Thank you very much for your help. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #302758 by </B><A HREF=/tutors/aboutme.mpl?userid=robertb><B>robertb(5830)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=robertb><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/cgi-bin/show-face.mpl?user=robertb><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=302758\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The degree of the denominator is greater than the degree of the numerator, and so for x-values close to infinity, the two dominant terms are related <BR>\n" );
document.write( "as <IMG STYLE=\"max-width:80%;\" SRC=/cgi-bin/plot-formula.mpl?expression=x%5E2%2Fx%5E3+=+1%2Fx ALIGN=MIDDLE  ALT=\"x%5E2%2Fx%5E3+=+1%2Fx\"   DISABLED_style= \"max-width:90%; height:auto;\" >, which would approach 0 as x goes to infinity.  The horizontal asymptote is thus y = 0. </SPAN>\n" );
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