limit of square root function $\sqrt{x-10}$ 
Evaluation of limits

$(a)\; \lim\limits_{x\to 10^{-}}\sqrt{x^2-100}$
$(b)\; \lim\limits_{x\to 7}\sqrt{x-10}$
$(a)$
Put $x=10-h$ and $\lim\limits_{x\to 10^{-}}=\lim_{h\to 0}$
Where $h=0.0000000000001$
$\lim\limits_{h\to 0}\sqrt{100+h^2-20h-100}=\sqrt{h}\sqrt{h-20}=0$
$(b)$
Put $\lim\limits_{x\to 7}\sqrt{x-10}=\sqrt{-3}=\sqrt{3}i$
Can anyone please tell me about my solution.
I have checked in Wolframalpha, it answers as $0$ and $\sqrt{3}i$
But I have seems that something wrong in $(a)$ and $(b)$ solution which I have given because of function domain.
For $1$ st $x\geq 10$ and for second $x\geq 10$
Thanks.
 A: If you are considering $\sqrt{x^2-100}$ and $\sqrt{x-10}$ as functions from $\Bbb R$ to $\Bbb R$, then these limits don't exist because the functions are undefined for $x<10$.
And you can't consider them as functions from $\Bbb C$ to $\Bbb C$, because the square root function is not unambiguously defined over $\Bbb C$.
So the only context that makes sense is if they are considered as functions from $\Bbb R$ to $\Bbb C$, which is rather an unlikely interpretation.
So I would say the question is badly posed.
A: Here you have pretty easy situation, because the limit is a value of function at given point (assuming that your domain is e.g. $\mathbb{C}$).
$(a)\; \lim_{x\rightarrow 10^{-}}\sqrt{x^2-100}=\sqrt{10^2-100}=\sqrt{0}=0$,
$(b)\; \lim_{x\rightarrow 7}\sqrt{x-10}=\sqrt{7-10}=\sqrt{-3}=\sqrt{3}i$.
You don't have to do nothing more than that.
Alternatively for second problem
Let $x:=7+h$, $h\rightarrow 0$. Then
$$\lim_{x\rightarrow 7}\sqrt{x-10}=\lim_{h\rightarrow 0}\sqrt{7+h-10}$$
which gives us the same result as above.
