How do you find $\lim _{x\to +\infty }\left(2^{1-\left(\frac{1}{2}\right)^x}\right)$ without using logarithms?

I tried this on Symbolab and here's what I got:

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However, I think I can do it this way: $$\lim _{x\ \rightarrow \ \infty \ }\ \left(2^{1-\left(\frac{1}{2}\right)^x}\right)=\left(\lim _{x\ \rightarrow \ \infty \ }2\ \right)^{\lim _{x\ \rightarrow \ \infty \ }\ \left(1-\left(\frac{1}{2}\right)^x\right)}=2^1=2.$$

I'm not sure if my method is accepted because I haven't seen any rules that state $$\lim _{x\rightarrow \ x_0\ }f\left(x\right)^{g\left(x\right)}=\lim _{x\ \rightarrow \ x_0\ }f\left(x\right)^{\lim _{x\ \rightarrow \ x_0\ }g\left(x\right)\ }$$ or at least $$\lim _{x\rightarrow \ x_0\ }c^{g\left(x\right)}=c^{\lim _{x\rightarrow \ x_0\ }g\left(x\right)\ }$$ with $c$ a given number and $\lim _{x\ \rightarrow \ x_0\ }f\left(x\right)$ and $\lim _{x\ \rightarrow \ x_0\ }g\left(x\right)$ exist.

Any help would be appreciated.

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    $\begingroup$ because we have no indeterminate forms and all the functions are continuous, we have $$\lim_{x\to\infty}2^{g(x}=2^{\lim_{x\to\infty}g(x)}=2^1$$ $\endgroup$ – marwalix Aug 10 at 10:42

As shown in this question, if $c,d$ are real numbers and $$\lim_{x\to x_0} f(x)=c>0, \quad\lim_{x \to x_0} g(x) =d>0,$$ then $$\lim_{x\rightarrow x_0}{f(x)^{g(x)}}=\left( \lim_{x\rightarrow x_0}{f(x)}\right)^{\left( \lim_{x\rightarrow x_0}{g(x)}\right)}=c^d,$$

so your original approach is correct. You may also recognize that

$$\lim _{x\to \infty }\left(2^{1-\left(\frac{1}{2}\right)^x}\right)=\lim _{x\to \infty }\left(\frac{2}{2^{{{(2^{-x})}}}}\right),$$

where provided $\lim\limits_{x \to x_0}f(x), \lim\limits_{x \to x_0}g(x)$ both exist and $\lim\limits_{x \to x_0}g(x)\neq 0$, you may apply the property

$$\lim_{x\rightarrow x_0}\left[\frac{f(x)}{g(x)}\right]=\frac{\lim\limits_{x \to x_0}f(x)}{\lim\limits_{x \to x_0}g(x)},$$


$$\lim _{x\to \infty }\left(2^{1-\left(\frac{1}{2}\right)^x}\right)=\lim _{x\to \infty }\left(\frac{2}{2^{{{(2^{-x})}}}}\right)=\frac{\lim\limits_{x \to \infty}2}{\lim\limits_{x \to \infty}2^{\left(\frac{1}{2^x}\right)}}=\frac{2}{2^0}=2.$$

In general, if the function is continuous then you may pass the limit operator inside and evaluate. As $2^{g(x)}$ is continuous for all $x$, we have

$$\lim _{x\to \infty }\left(2^{1-\left(\frac{1}{2}\right)^x}\right)=2^{\lim\limits_{x \to \infty}1-\left(\frac{1}{2}\right)^x}=2^{1-0}=2.$$

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The function $g(y) = 2^y$ is continuous. So, the limit is equal to $2^{\lim_{x \rightarrow \infty} 1 - \frac{1}{2^x}} = 2$

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  • $\begingroup$ +1 Thank you for your answer. I have one more question: Are all functions of the form $a^x$ with $a \in \mathbb{R}$ continuous on $\mathbb{R}$? $\endgroup$ – Dave Robin Aug 10 at 13:06
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    $\begingroup$ As long as a > 0. Note that if a < 0, a^{1/2} is not even defined. If a = 0, 0^b is not defined for b \leq 0. $\endgroup$ – abcd123 Aug 10 at 13:21

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