Limits to infinity

Question

lim as x tends to infinity of (1/5)ln(x/2x+5) 

 

The answer is -ln(2/5) . I dont understand how you get this. 

Answer 1

Assuming the correct expression is

 

(1/5) ln[x / (2x + 5)],

 

examine the argument of the ln function as x→∞.

 

x / (2x + 5) → x/(2x) = 1/2

 

Then

 

(1/5) ln[x / (2x + 5)] → (1/5) ln(1/2) = -(1/5)ln2

 

This is the same answer Kenneth got, so I have to agree with him that there is an error in the problem statement.

 

 

Answer 2

I'm pretty sure that the denominator (2x+5) also REQUIRES parenthesization.

 

simplify ln [1/(2+5/x] and then as x goes to infinity, you've got this limit,m as approaches infinity:

ln(1/2) = ln 1 - ln 2 = - ln2.

 

Putting it all together, I get (1/5)(-ln2) BUT THIS IS NOT WHAT YOU WROTE AS THE SUPPOSED ANSWER. (I think that your parenthesization is in error.)