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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 $\lim\limits_{x\to1} \left(\frac {1+x} {2+x}\right)^{\frac {1-\sqrt x} {1-x}}$?

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    Is it supposed to be $x\to 1$ instead of $n\to 1$? – 2017-01-24
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    Hint: $1-x = (1-\sqrt{x})(1+\sqrt{x})$. – 2017-01-24
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    @ajotatxe Yes, sorry. Gonna correct it. – 2017-01-24
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    $1-x=(1-\sqrt{x})(1+\sqrt{x})$ – 2017-01-24
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    @ReinhardMeier Okay it seems easy applying that, but why am I allowed to say that $\frac x x = 1$ when $x = 0$? – 2017-01-24
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    Recall the definition of the limit. The $\frac{x}{x}$ does not have to be defined at $x=0$ in order to allow the calculation of $\lim_{x\rightarrow 0}\frac{x}{x}$. $\frac{x}{x}$ only needs to approach a value, when $x$ gets closer to $0$. – 2017-01-24
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3 Answers 3

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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 $f(1) = \lim_1 f$.

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    Mmh. Why the downvote, if I may? (Is there something wrong in my answer, and if so what?) – 2017-01-24
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Note that for $x\neq 1$, $$f(x)=\left(1-\frac{1}{2+x}\right)^{\frac{1}{1+\sqrt x}}$$ so $$\lim_{x\to 1}f(x)=\left(\frac23\right)^{1/2}=\sqrt\frac23$$

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    I was taught that cancellation is not allowed for variable that are 0 why are you allowed to do it here? – 2017-01-24
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    @user405981 Because $x\not=1$ or differently the denominator is never zero,close to zero but never zero. – 2017-01-24
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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

$$\lim_{x\to1}\frac{1-x}{1-\sqrt x}=\lim_{x\to1}\frac1{1+\sqrt x}=\frac12.$$