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Mathematics 16 Online
OpenStudy (anonymous):

limit with x approaching infinity at (2x -1)/(|x| - 3)

OpenStudy (anonymous):

l'hopital's rule

OpenStudy (anonymous):

can you show me with and without hopitals plz

OpenStudy (anonymous):

I don't think that's necessary, though you can certainly do that. \(|x|=x\) for positive \(x\), which is the case since \(x\to\infty\).

OpenStudy (anonymous):

yeahbut i still dont get it

OpenStudy (anonymous):

\[\lim_{x\to\infty}\frac{2x-1}{|x|+3}=\lim_{x\to\infty}\frac{2x-1}{x+3}=\lim_{x\to\infty}\frac{2-\frac{1}{x}}{1+\frac{3}{x}}=\cdots\]

OpenStudy (anonymous):

2?

OpenStudy (anonymous):

Sorry, minus in the denominator, not plus. The answer doesn't change anyway.

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

thank you

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