Calculate $\lim\limits_{x\to1} (\frac {1+x} {2+x})^{\frac {1-\sqrt x} {1-x}}$ I've drawn the graph of the function $f$ defined by $f(x) = \left(\frac {1+x} {2+x}\right)^{\frac {1-\sqrt x} {1-x}}$ by calculating as many points as possible and draw lines between them. So I saw that $f$ probably is continuous at 1. However drawing the graph does not mean proving my assumption.
So I tried to calculate $\lim\limits_{x\to1}f(x)$, because if I am not mistaking proving $\lim\limits_{x\to1}f(x) = c$ with $c$ being any constant in $\mathbb R$ means that the graph is continuous at 1. However, I don't know a way to do that. I know the L'Hopital rule but I wasn't taught that there would be a way to apply it to exponents. 
Is there a rule to calculate $\lim\limits_{x\to1} \left(\frac {1+x} {2+x}\right)^{\frac {1-\sqrt x} {1-x}}$?
 A: As usual, having $x$ in both the base and the exponent is confusing. Start by rewriting it in its exponential form, namely
$$
f(x) = \left(\frac {1+x} {2+x}\right)^{\frac {1-\sqrt x} {1-x}}
= \exp\left(\frac {1-\sqrt x} {1-x} \ln \frac {1+x} {2+x}\right) \tag{1}
$$
Now, that clears it up a little: by continuity of the exponential, we only have to try to analyze the behavior of $\frac {1-\sqrt x} {1-x} \ln \frac {1+x} {2+x}$ when $x\to 1$. Since 
$
\frac {1+x} {2+x} \xrightarrow[x\to 1]{} \frac{2}{3}
$
and $\ln$ is continuous, we have $$\ln \frac {1+x} {2+x}\xrightarrow[x\to 1]{} \ln\frac{2}{3} \tag{2}$$ so we only have to handle now the first factor, $\frac {1-\sqrt x} {1-x}$. 
For $x>0$ (which is surely the case when $x$ is close to $1$) different than $1$,
$$
\frac {1-\sqrt x} {1-x} = \frac {1-\sqrt{x}} {1^2-\sqrt{x}^2}
= \frac {1-\sqrt{x}} {(1-\sqrt{x})(1+\sqrt{x})}
= \frac {1} {1+\sqrt{x}} \xrightarrow[x\to 1]{} \frac{1}{2} \tag{3}
$$
which solves this part.
Putting it together,
$$
f(x) =  \exp\left(\frac {1-\sqrt x} {1-x} \ln \frac {1+x} {2+x}\right) \xrightarrow[x\to 1]{} \exp\left( \frac{1}{2}\ln\frac{2}{3}\right) 
= \boxed{\sqrt{\frac{2}{3}}}.
$$

Regarding the rest of your question: having that $\lim_1 f$ exists in $\mathbb{R}$ does not guarantee that $f$ is continuous at $1$, it guarantees that $f$ can be extended by continuity at $1$. The distinction in your case is mostly of vocabulary, and sounds a bit strange, but imagine the following: I could define $f(1) = 97$ if I pleased. $\lim_1 f$ would still exist and be equal to $\sqrt{\frac{2}{3}}$, but the function $f$ would now be discontinuous at $1$. To have it continuous, you also need to set $f(1) = \lim_1 f$.
A: Note that for $x\neq 1$,
$$f(x)=\left(1-\frac{1}{2+x}\right)^{\frac{1}{1+\sqrt x}}$$
so
$$\lim_{x\to 1}f(x)=\left(\frac23\right)^{1/2}=\sqrt\frac23$$
A: Using $\lim a^b=(\lim a)^{\lim b}$ when the limits are defined, we have
$$\lim_{x\to1}\left(\frac {1+x} {2+x}\right)^{\frac {1-\sqrt x} {1-x}}=\left(\frac23\right)^{\lim_{x\to1}(1-x)/(1-\sqrt x)}.$$
Then
$$\lim_{x\to1}\frac{1-x}{1-\sqrt x}=\lim_{x\to1}\frac1{1+\sqrt x}=\frac12.$$
