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is it true that : $$\lim_{x\to +\infty} 1^{x} = 1$$

because we can write this limit as : $$ \lim_{x\to +\infty} e^{~x\ln(1)}$$

but this limit is an indertminate form : $$\lim_{x\to+\infty} x\ln(1)$$

so how to evaluate this limit ?

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    Remember that $\log 1 = 0$. – 2017-01-31
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    Hwo do **you define** $\;1^x\;$ when, for example, $\;x=\frac12\,,\,x=\pi\;$ , etc.? If you define it by means of the exponential then all is easy, since $\;\log1=0\;$ , so you can multiply by any **number** and it still is zero... – 2017-01-31

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One may observe that $$ 1^x=e^{~x\ln(1)}=e^{~x\cdot0}=e^0=1,\quad x \in \mathbb{R}, $$ giving $$ \lim_{x \to \infty}1^x=\lim_{x \to \infty}1=1. $$

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$\lim \limits_{x \to + \infty} x \ln 1$ is not indetermined.

Since $x \ln 1 = x ~0 = 0 ~\forall x$ also $\lim \limits_{x \to + \infty} x \ln 1 = 0$

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    but this limit is equivalent to $\infty . 0$, which is indetmined. – 2017-01-31
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    The form might be indetermined, but it does not mean in general that it is undoable. For example, $\lim_{x\to \infty} 1 = 1$, but we can also make this indetermined by writing $\lim_{x\to \infty} 1 = \lim_{x\to \infty} x (1/x)$. @Hilbert – 2017-01-31
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Hint:Any limit of the form $\lim_{x\to\infty}{f(x)^{g(x)}}$ where,for the given limit $$f(x)\rightarrow1 \\ g(x)\rightarrow+\infty$$ equals to $e^{\lim_{x\to\infty}{(f(x)-1)\cdot g(x)}}$

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You don't need to use any algebra, just directly compute the limit:

$$\forall \epsilon >0, \exists c, \forall x > c : | 1^x - 1 | < \epsilon $$

The thing is, you can let $x$ be any positive number, because $1^x = 1$ for any $x > 0$; so choose $c = 0$.