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Show that the function $ f:(−1,0)∪(0,1) → R , x → { 1 \over {1 \over x} +1} $ converges at $ a = 0$

The function ${ 1 \over {1 \over x} +1}$ is defined on the interval $ f:(−1,0)∪(0,1) $ excluding the point $x = 0$ because the denominator $ {1 \over x} $ causes the function to be undefined at $x = 0$.

$ { 1 \over {1 \over x} +1} = { x \over 1+x} $

Let's compute the limit of the function $ f(x) $ as $ x $ approaches to zero:

$ lim_{x \to 0} f(x) = lim_{x \to 0} { x \over 1 + x} = { 0 \over 1 +0 } = 0$

Hence, the limit of the function ${ 1 \over {1 \over x} +1} $ as $x$ approaches to $0$ is $0$. This signifies that the function converges to $0$ if $x$ converges to $0$.

Is it correct?

I don't need a name
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  • Looks fine to me. Another approach is to note that $g(x) \mapsto {x \over 1+x}$ is continuous on $(-1,1)$ and $f$ equals $g$ on the domain of $f$. – copper.hat Nov 15 '23 at 04:51

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