Does the limit of this function exist? If so what is its value? x^2+2x-6 / 3x^2+2
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HallsofIvy
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In other words, do you mean \(\displaystyle \lim_{x\to \infty} x^2+ 2x- \frac{6}{3x^2}+ 2\), which is what you wrote, or do you mean \(\displaystyle \lim_{x\to\infty}\frac{x^2+ 2x- 6}{3x^2+ 2}\), which is more likely?
For x very very large, the highest power of x "dominates". For example, with x= 100, \(\displaystyle 3x^2+ 2= 30000+ 2= 30002\) and if x= 1000, \(\displaystyle x^2+ 2= 3000000+ 2= 3000002\).
For x very very large, the highest power of x "dominates". For example, with x= 100, \(\displaystyle 3x^2+ 2= 30000+ 2= 30002\) and if x= 1000, \(\displaystyle x^2+ 2= 3000000+ 2= 3000002\).
Dr.Peterson
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Assuming you meant lim{x->infty} (x^2+2x-6) / (3x^2+2), a standard way to demonstrate such a limit is to divide the numerator and the denominator by the highest power of x, namely x^2. Do you see what happens to the numerator and denominator when you then let x approach infinity?