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$f(x)=x-(x+k)^\epsilon$ where $k>1$ and $0<\epsilon<1$.

How do I show that $f(x)=0$ has a root at $x>0$?

The only one root part is obvious, but I am asking this to rigorously show that there is a root. I have an argument involving the growth rates of $x$ and $(x+k)^\epsilon$ in my mind, but somehow I feel that is not rigorous enough.

The problem came up in another problem involving some limits, but this is the only step prevening me from reaching the solution. Thanks in advance.

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Here a proof that a root exists: first, note that $f(1)<0$. Then, note that $$ \lim_{x\to \infty}\frac{x}{(x+k)^{\epsilon}} > 1 $$ So, there exists a $c>0$ with $c>(c+k)^{\epsilon}$. By the intermediate value theorem, $f$ has a root.

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