Infinite Limit Question

Feb 2010
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2
Can anyone please explain to me how \(\displaystyle \lim_{n->\infty}\left| \frac{nx}{n+1} \right| = |x|\)? Does it not require L' Hopital's Rule and if so, how does it evaluate to |x|?
 

TheEmptySet

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Feb 2008
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Can anyone please explain to me how \(\displaystyle \lim_{n->\infty}\left| \frac{nx}{n+1} \right| = |x|\)? Does it not require L' Hopital's Rule and if so, how does it evaluate to |x|?
\(\displaystyle \left| \frac{nx}{n+1} \right|=\bigg|\frac{n}{n} \bigg|\left| \frac{x}{1+\frac{1}{n}} \right| = \left| \frac{x}{1+\frac{1}{n}} \right| \)
 
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matheagle

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Feb 2009
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I'd pull out the |x| since n is certainly positive.

\(\displaystyle \left | \frac{nx}{n+1} \right| = |x|\left({n\over n+1}\right)\to |x|\cdot 1=|x|\)