Limit as x-> infinity
- Thread starter JSmith
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HallsofIvy
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Do you understand why that is true? Dividing both numerator and denominator of
\(\displaystyle \frac{-3x^2+ 6x- 4}{2x^2- 5}\)
by \(\displaystyle x^2\) gives
\(\displaystyle \frac{-3+ \frac{6}{x}- \frac{4}{x^2}}{2- \frac{5}{x^2}}\)
and, of course, as x goes to infinity, those fractions all go to 0 leaving -3/2.
Similarly, dividing both numerator and denominator of
\(\displaystyle \frac{2x- 11}{2x^2+ 2x- 1}\)
by \(\displaystyle x^2\) gives
\(\displaystyle \frac{\frac{2}{x}- \frac{11}{x^2}}{2+ \frac{2}{x}- \frac{1}{x^2}}\)
and, again, those fractions each go to 0, leaving 0/2= 0.
\(\displaystyle \frac{-3x^2+ 6x- 4}{2x^2- 5}\)
by \(\displaystyle x^2\) gives
\(\displaystyle \frac{-3+ \frac{6}{x}- \frac{4}{x^2}}{2- \frac{5}{x^2}}\)
and, of course, as x goes to infinity, those fractions all go to 0 leaving -3/2.
Similarly, dividing both numerator and denominator of
\(\displaystyle \frac{2x- 11}{2x^2+ 2x- 1}\)
by \(\displaystyle x^2\) gives
\(\displaystyle \frac{\frac{2}{x}- \frac{11}{x^2}}{2+ \frac{2}{x}- \frac{1}{x^2}}\)
and, again, those fractions each go to 0, leaving 0/2= 0.