Question

Find the limit as x approaches positive infinity of [2x/(3x+1)]-[(x^2)/(x-3)]

Answer 1

familiar with l'Hopital's Rule?

Answer 2

...no?

Answer 3

actually don't use it.

re-arrange it like this:

\[\frac{ 2x }{ 3x + 1 } - \frac{ x^2 }{ x-3 } = \frac{ x }{ x }\frac{ 2 }{ 3 + \frac{ 1 }{ x } } - \frac{ x }{ x }\frac{ x }{ 1 -\frac{ 3 }{ x } } = \frac{ 2 }{ 3 + \frac{ 1 }{ x } } - \frac{ x }{ 1 - \frac{ 3 }{ x} }\]

\[\lim_{x->\infty}\left( \frac{ 2 }{ 3 + \frac{ 1 }{ x } } - \frac{ x }{ 1 - \frac{ 3 }{ x} } \right) = \frac{ 2 }{ 3 } - \frac{ \infty }{ 1 } = -\infty\]

Answer 4

Thank you so much! I still might look up l'Hopital's Rule anyways...

Answer 5

it is very useful and typically obtains results faster than this. you will definitely eventually learn it in your calculus course but being ahead is always better :)