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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?

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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 $$\left(\left(1+\frac{1}{2^x}\right)^{2^x} \right)^{x/2^x}.$$

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  • $\begingroup$ I see. The limit of x/(2^x) can be evaluated with l'hospital's rule to 1/ln(2)*2^x, which evaluates to 1/inf and therefore 0, so that makes sense. I'm still thinking on how to use squeezing and continuity for this problem. $\endgroup$ – Topher Mar 11 '15 at 22:55
  • $\begingroup$ Oh, that second approach makes sense....ugh... :D $\endgroup$ – Topher Mar 11 '15 at 22:57
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    $\begingroup$ The second approach is virtuous, since it recycles the fact that $\lim_{y\to\nfty}(1+1/y)^y=e$, although really we only need boundedness, since the outer exponent $x/2^x$ goes to $0$. In the first approach, since $\frac{x}{2^x}\to 0$, the ln of $(1+1/2^x)^x$ is squeezed between $0$ and something that goes to $0$, so it has limit $0$. And since the ln of our expression has limit $0$, our expression has limit $e^0$. That's because of the continuity of the exponential function. $\endgroup$ – André Nicolas Mar 12 '15 at 0:08
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$$\Large \lim_{x\to \infty} \left( 1+\frac{1}{2}^{x} \right)^{x}=\lim e^{x\log (1+1/2^x)}=e^{\lim {x\log (1+1/2^x)}}=e^{\lim_{t\to 0} \log (1+1/2^{1/t})/t}=e^0=1$$

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  • $\begingroup$ This makes some sense, but by inverting the variable, aren't you still left with the limit being in the 0/0 indeterminate form, as opposed to inf/inf in the original form? In essence, this still does not give you a determinate way to get the value of the limit. You are still left in a "race" between the numerator and denominator in the limit. $\endgroup$ – Topher Mar 11 '15 at 22:46
  • $\begingroup$ In the original form, it was $1^\infty$. In the 3rd one, it is $\infty\times\infty$. I will add more steps on how to calculate the limit in the 4th step. $\endgroup$ – Cyclohexanol. Mar 11 '15 at 23:11
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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.

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  • $\begingroup$ $1^\infty \neq 1$, ex $$\lim (1+1/n)^n=1^\infty=e$$ $\endgroup$ – Cyclohexanol. Mar 11 '15 at 23:09
  • $\begingroup$ Ugh...ok, I'm kind of stuck in this understanding. Don't 1^1, 1^2, 1^3,...1^n=1, so wouldn't 1^n as n-->inf also be 1? $\endgroup$ – Topher Mar 11 '15 at 23:21
  • $\begingroup$ Not as a composition of functions. For example, $\lim 1^n=1$, but $\lim (1+3^{-n})^n\neq 1$ $\endgroup$ – Cyclohexanol. Mar 11 '15 at 23:25

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