Determine limn→∞n(1+(n+1)lnnn+1)
I noticed the indeterminate case ∞⋅0 and I tried to get them all under the ln, but it got more complicated and I reached another indeterminate form. The same happened when I tried to use Stolz-Cesaro.
EDIT: is there an elementary solution, without l'Hospital or Taylor series?
Answer
Let nn+1=ex. As n→∞,x→0 through negative values.The expression is ex⋅1+x−ex(1−ex)2.
Applying l'Hopital's Rule to 1+x−ex(1−ex)2 we get a limit of −12 as x→0. And the far left term ex in the expression goes to 1 as x→0.
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