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Sequences and Limits

  • Thread starter cauchy21
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(a) [tex]lim_{n\rightarrow\infty}[/tex] ([tex]\sqrt{(n + a)(n + b)} - n)[/tex] where a, b > 0
(b)[tex]lim _{n\rightarrow\infty}[/tex] (n!)1/n2
 
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Answers and Replies

  • #2
HallsofIvy
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(a) [tex]lim_{n\rightarrow\infty}[/tex] ([tex]\sqrt{(n + a)(n + b)} - n)[/tex] where a, b > 0
Rationalize the numerator by multiplying both numerator and denominator by [itex]\sqrt{n+ a)(n+b)}+ n[/itex]. Then divide both numerator and denominator by n.

[/quote](b)[tex]lim _{n\rightarrow\infty}[/tex] (n!)1/n2[/QUOTE]
If [itex]y= (n!)^{1/n^2}[/itex] then
[tex]ln(n)= \frac{ln(n!)}{n^2}= \frac{ln(2)}{n^2}+ \frac{ln(3)}{n^2}+ \cdot\cdot\cdot+ \frac{ln(n)}{n^2}[/tex].
 

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