I have the equation:
$y=\left( 1+\frac{1}{2}^{x} \right)^{x}$
In evaluating it's limit as it approaches +$\infty$, I can't seem to simplify the expression to a non-indeterminate form. By graphing I know the limit is 1, but I'd like to show this mathematically. At ever step, be it taking the natural log and then using L'Hospital's rule, or otherwise, I end up with another indeterminate form of the limit.
Any suggestions on how to tackle this?
On
$$\Large \lim_{x\to \infty} \left( 1+\frac{1}{2}^{x} \right)^{x}=\lim e^{x\log (1+1/2^x)}=e^{\lim {x\log (1+1/2^x)}}=e^{\lim_{t\to 0} \log (1+1/2^{1/t})/t}=e^0=1$$
On
In re-visiting this, I think I overlooked something simple. I was bent on applying l'hospital's rule when this wasn't even required:
$$\lim_{x\to \infty} \left( 1+\frac{1}{2}^{x} \right)^{x}=\lim_{x\to \infty} \left( 1+\frac{1^{x}}{2^{x}} \right)^{x}=\lim_{x\to \infty} \left( 1+\frac{1}{2^{x}} \right)^{x}$$
This can be evaluated directly, with no additional conversion:
$$\lim_{x\to \infty} \left( 1+\frac{1}{2^{x}} \right)^{x}=\left( 1+\frac{1}{2^{\infty}} \right)^{\infty}=\left( 1+0 \right)^{\infty}=1$$
...unless this is somehow not a proper form of showing or proving this evaluation, but it seems to make sense to me.
Note that if $t\gt 0$ then $\ln(1+t)\lt t$. The logarithm of our expression is $x\ln\left(1+\frac{1}{2^x}\right)$. Thus $$0\lt x\ln\left(1+\frac{1}{2^x}\right)\lt \frac{x}{2^x}.$$ It is a familiar fact that $\lim_{x\to\infty}\frac{x}{2^x}=0$. The rest follows by Squeezing and continuity.
Another approach: Our expression is equal to $$\left(\left(1+\frac{1}{2^x}\right)^{2^x} \right)^{x/2^x}.$$