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I have to prove that the following function is continuous and I have to find a fixed point:

$f: \mathbb{R}\rightarrow\mathbb{R}, x \rightarrow\frac{(n-1)x^n+a}{nx^{n-1}}$ with $n \in \mathbb{N}$ and $a \in \mathbb{R_\geq0}$

We defined that a function is continuous if for a sequence $(x_n)\in D$ with $f: D \rightarrow E$ with lim $(x_n)$ = $\delta$ it follows that lim$f(x_n)$ = $f(\delta)$

I could prove my task with the epsilon delta criterion but our definition gives me headache.

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    The function is not well-defined at $x=0$. – 2017-01-22

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Define $f$ for $x\ne 0$. Then $f$ is continuous as a rational function. To find a fixed point it os enough to solve the equation $f(x)=x$, which leads to $x^n=a$, so $x=\sqrt[n]{a}$ is a fixed point.

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    How did you solve it? I get a different solution for x – 2017-01-22
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    $\dfrac{(n-1)x^n+a}{nx^{n-1}}=x\iff (n-1)x^n+a=nx^n\iff x^n=a$ – 2017-01-22