Help with the limit
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Jan 22, 2011 #2 Hint: ln(n) <= n and for n>2, 1 < ln(n)
K koutos New member Joined Jan 22, 2011 Messages 2 Jan 22, 2011 #3 Thanks for the hint, it was very useful. I solved it, i think. Here is my solution: Am i right?
D daon Senior Member Joined Jan 27, 2006 Messages 1,284 Jan 22, 2011 #4 you got it! I would make mention of the squeeze theorem. Words in proofs aren't bad things
L lookagain Elite Member Joined Aug 22, 2010 Messages 3,255 Jan 22, 2011 #5 koutos said: Am i right? Click to expand... I see the following to finish part of the last two lines: 1n≤ln(n)n≤nn\displaystyle \sqrt[n]{1} \le \sqrt[n]{ln(n)} \le \sqrt[n]{n}n1≤nln(n)≤nn limn→∞1n=limn→∞(n)n→limn→∞ln(n)n=1\displaystyle \lim_{n \to \infty}\sqrt[n]{1} = \lim_{n \to \infty}\sqrt[n]{(n)} \rightarrow \lim_{n \to \infty}\sqrt[n]{ln(n)} = 1n→∞limn1=n→∞limn(n)→n→∞limnln(n)=1 The middle expression in your bottom line of this quote box looks wrong to me.
koutos said: Am i right? Click to expand... I see the following to finish part of the last two lines: 1n≤ln(n)n≤nn\displaystyle \sqrt[n]{1} \le \sqrt[n]{ln(n)} \le \sqrt[n]{n}n1≤nln(n)≤nn limn→∞1n=limn→∞(n)n→limn→∞ln(n)n=1\displaystyle \lim_{n \to \infty}\sqrt[n]{1} = \lim_{n \to \infty}\sqrt[n]{(n)} \rightarrow \lim_{n \to \infty}\sqrt[n]{ln(n)} = 1n→∞limn1=n→∞limn(n)→n→∞limnln(n)=1 The middle expression in your bottom line of this quote box looks wrong to me.
