Can anyone supply an answer to this apparent simple contradiction? (Sorry but I have written X.X for X squared as I have no power facility on this network) Consider y = (X.X + 3X + 2 )/(X-1) = (X + 3 + 2/X)/(1-1/X) as X tends to infinity 1/X tends to zero and y tends to X + 3. But y = X + 4 + 6/(X-1) and as X tends to infinity y tends to X + 4. There is obviousy no problem with division by zero as X is tending to infinity. For large X, y does tend to X + 4. This can be checked with say X = 1000. y then becomes 1004.006006 Bob Hooper
Michael Fisher | (203.88.235.50) | Thursday, 18 April 2002 11:40:33 PM
Consider X = ten million. Y = 10,000,000 + 3 + .0000002 divided by .9999999 which is pretty close to 10,000,003. I don't know how you derived the second solution to be y=X+4+6/(X-1), but the first equation does tend to X+3 as X approaches infinity. At only 1,000 there is enough other stuff still there to boost it over the line to 1,004. For large X though, Y tends to X+3.
John Smith | (203.166.2.146) | Thursday, 13 June 2002 11:23:03 AM
the solution you got by long division (xx+3x+2)/(x-1)=x+4+6/(x-1) giving y=x+4 is correct as can be seen by considering |(x+4)-(x+4+6/(x-1)| which tends to 0 as x tends to infinity. Similarly y=x+3 is incorrect since |x+3-(xx+3x+2)/(x-1)| = (x+5)/(x+1) which tends to 1 as x tends to infinity. I think that all you can say is that you cannot get the oblique asymptote by dividing top and bottom by x. Why should you?
james lim | (203.131.64.140) | Sunday, 7 July 2002 1:39:51 PM
i need help. how can we determine oblique asymptotes? could you help me?
jay | (161.114.88.71) | Wednesday, 25 September 2002 2:55:02 AM
what is the vertical asymptote for f(x)=5x+7/x2-10x+24 and what is the oblique asymptote for f(x)=2x3+9x2+11x+10/x2+4x+3. The numbers after the x are exponents. I need some help please.
jay | (161.114.88.71) | Wednesday, 25 September 2002 2:56:19 AM
what is the vertical asymptote for f(x)=5x+7/x2-10x+24 and what is the oblique asymptote for f(x)=2x3+9x2+11x+10/x2+4x+3. The numbers after the x are exponents. I need some help please.
jay | (161.114.88.71) | Wednesday, 25 September 2002 2:57:17 AM
what is the vertical asymptote for f(x)=5x+7/x2-10x+24 and what is the oblique asymptote for f(x)=2x3+9x2+11x+10/x2+4x+3. The numbers after the x are exponents. I need some help please.
Johnny | (12.246.237.156) | Thursday, 22 May 2003 3:29:56 PM
Wow this is old!
lalala | (203.131.147.185) | Saturday, 13 December 2003 2:55:57 PM
oh my!