simple limit question

bomba923

How do I show that [tex] \mathop {\lim }\limits_{n \to \infty } \left( {n!!} \right)^{\left( {n^{ - n} } \right)} = 1 [/tex] ?

(The n^-n forces the value to decrease faster than n!! increases, I believe. But how to work out that?)

estalniath

1)Firstly we'll let the expression (n!!)^(n^-n)=y, then taking ln on both sides give,
lny=ln(n!!)/n^n.
2) We'll then find the limit of the expression at the RHS using Le Hopital's rule or by observation that n^n increases faster than ln(n!!). =)
3) After finding the limit, L, all we have to do is substitute back the value of y, which is e^L

Similar Threads for: simple limit question
Thread	Forum	Replies
simple limit question. need a little help please	Calculus	7
simple question on limit	Calculus & Beyond Homework	3
Simple limit question.	Calculus & Beyond Homework	7
Simple Limit question	General Math	6
Simple limit question	Introductory Physics Homework	7

bomba923	simple limit question Tweet How do I show that [tex] \mathop {\lim }\limits_{n \to \infty } \left( {n!!} \right)^{\left( {n^{ - n} } \right)} = 1 [/tex] ? (The n^-n forces the value to decrease faster than n!! increases, I believe. But how to work out that?)

estalniath	1)Firstly we'll let the expression (n!!)^(n^-n)=y, then taking ln on both sides give, lny=ln(n!!)/n^n. 2) We'll then find the limit of the expression at the RHS using Le Hopital's rule or by observation that n^n increases faster than ln(n!!). =) 3) After finding the limit, L, all we have to do is substitute back the value of y, which is e^L