1. Feb 11, 2009 #1

   transgalactic

   

   [tex]
   lim_{x->\infty}\frac{3x-cosx}{4x+sinx}=lim_{x->\infty}\frac{\frac{3x-cosx}{x}}{\frac{4x+sinx}{x}}=lim_{x->\infty}\frac{3-0}{4-0}=0.75
   [/tex]
   
   i thought bounded/infinity=0
   
   
   ??

    

   transgalactic, Feb 11, 2009
2. 
   Phys.org - latest science and technology news stories on Phys.org

   • • Scarlet macaw DNA points to ancient breeding operation in Southwest
   • • Mathematicians solve age-old spaghetti mystery
   • • New technology can detect hundreds of proteins in a single sample
3. Feb 12, 2009 #2

   Mark44

   

   Insights Author

   Staff: Mentor

   You have the correct limit, but you should remove the limit operator from the expression below, since you have already "passed to the limit" in the previous expression:
   [tex]lim_{x->\infty}\frac{3-0}{4-0}=0.75 [/tex]

    

   Mark44, Feb 12, 2009
4. Feb 12, 2009 #3

   transgalactic

   

   thanks

    

   transgalactic, Feb 12, 2009

• Log in with Facebook
• Log in with Twitter

Your name or email address:

Do you already have an account?

• No, create an account now.
• Yes, my password is:
• Forgot your password?

Stay logged in