Can you say that the function $ y(x)=1/(1/x)$ admits no solution for $x = 0$? 
Can you say that the function $ y(x)=1/(1/x)$ admits no solution for $x = 0$?

Or is that function too "close" to y(x)=x to say that? 
 A: As written the function has no value at $x=0$. It does however admit a limit: $\lim_{x\rightarrow 0}f(x)=0$ so it can be extended to a continuous function $g(x):=f(x)$ for $x\neq 0$ and $g(0)=0$. In other words, $g(x)=x$. 
A: It admits no value as it is not defined at $x=0$!
A: The answer is Yes.  You must be able to plug your solution back into the original equation.  Taking limits is not a solution, since limits don't care what's going on at x = 0, instead limits look at what is happening in the neighborhood of x = 0.  
A: I will provide a similar example to explain what happens. Let $f(x)=x-1$.
Then $x=1$ is the unique solution to the equation $f(x)=0.$ In other words,
the function $f$ have exactly one root which is $x=1.$ Note that this root
is an element of the domain of definition of the function $f$ which is by
the way the real set $%
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\mathbb{R}
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$ since $f$ is polynomial. 
Now, consider $g(x)=\frac{x^{2}-1}{x+1}.$ Its domain is $
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\mathbb{R}
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-\{-1\},$ so when one want to search roots of $g$ we should for them inside
the domain of $g.$ That is, the element $x=-1$ is systematical excluded
first (before starting looking for roots to $g).$ Next we solve the equation 
$g(x)=0,$ by considering $x$ inside $Dom(g),$ so $x\neq -1,$ then we write
\begin{eqnarray*}
\frac{x^{2}-1}{x+1} &=&\frac{(x-1)(x+1)}{(x+1)},\ for\ x\in 
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%BeginExpansion
\mathbb{R}
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-\{-1\} \\
&=&(x-1),\ \ \ \ \ for\ x\in 
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\mathbb{R}
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-\{-1\}
\end{eqnarray*} then we solve 
\begin{eqnarray*}
x-1 &=&0 \\
x &=&1
\end{eqnarray*}Now before before stating that there is exactly one root for $g$ which is $%
x=1,$ we check if this 'candidate' is inside the $dom(g).$ After that, we
say okay, because $x=1$ is inside the domain then it is in the solution-set.
Since there is no other one, it consists of the whole solution set..
Note that $g(x)=f(x)$ for $x\in 
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\mathbb{R}
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-\{-1\}$
