Limit of exponential fcn: [( 2 ^ x ) + 1] / [ 2^ ( x + 1) ]

[peblez]

lim x--> infinity [( 2 ^ x ) + 1] / [ 2^ ( x + 1) ]

THis is what i did. I don't know exactly how to solve for this but , my logic is that it'll go to zero eventually, because the bottom goes to infinity faster than the top so it'll to go zero. But i am not sure exactly if this is right can someone prove this or tell me if i'm intuatively correct. thanks

 

[daon]

Re: Limit of a exponential function

lim x--> infinity [( 2 ^ x ) + 1] / [ 2^ ( x + 1) ]

THis is what i did. I don't know exactly how to solve for this but , my logic is that it'll go to zero eventually, because the bottom goes to infinity faster than the top so it'll to go zero. But i am not sure exactly if this is right can someone prove this or tell me if i'm intuatively correct. thanks

\(\displaystyle \L \frac{2^x+1}{2^{x+1}} = \frac{1}{2}+\frac{1}{2^{x+1}\)

The second term approaches zero, but not the first. Since the limit of each term exists, use the fact that the limit of the sum is the sum of the limits. So it actually approaches 1/2, not zero.

 

[soroban]

Re: Limit of exponential fcn: [( 2 ^ x ) + 1] / [ 2^ ( x + 1

Hello, peblez!

Another approach . . .

\(\displaystyle \L\lim_{x\to\infty}\,\frac{2^x\,+\,1}{2^{x+1}}\)

Divide top and bottom by \(\displaystyle 2^x\)

. . \(\displaystyle \L\lim_{x\to\infty}\,\frac{\left(\frac{2^x\,+\,1}{2^x}\right)}{\left(\frac{2^{x+1}}{2^x}\right)} \;=\;\lim_{x\to\infty}\,\frac{1\,+\,\frac{1}{2^x}}{2} \;=\;\frac{1\,+\,0}{2}\;=\;\frac{1}{2}\)

 

[peblez]

Ah. Thank you.