I know that $$\lim_{x\to 0} \sqrt{x}$$ does not exist when $x\to 0$. By looking at the graph of $\sqrt{x}$ it is obvious that yes it is not defined for any negative value of $x$. But as I have been taught that for any limit to exist it must satisfy:
$$\lim_{x\to 0^+} \sqrt{x}=\lim_{x\to 0^-} \sqrt{x}$$
So when I tried to prove it by subsituting $x=a+h$ for RHL & $x=a-h $ for LHL, where $a=0$ & $h\to 0$
RHL$$=\lim_{h\to 0^+} \sqrt{a+h}$$
$$=\lim_{h\to 0^+} \sqrt{0+h}$$
$$=\lim_{h\to 0^+} \sqrt{h}=0$$ AND
LHL$$=\lim_{h\to 0^-} \sqrt{a-h}$$
$$=\lim_{h\to 0^-} \sqrt{0-h}$$
$$=\lim_{h\to 0^-} \sqrt{-h}=0$$
So RHL $=$LHL. Says that limit exists but we all know that's not the answer. So what's the problem for this method?
My second question is that I have seen multiple times for questions like
$$\lim_{x\to6} \dfrac{(x+6)(x-6)}{\sqrt{(x-6)^2}}$$, which simplifies to $x+6$ if denominator's square root is open. But this opening of square root is not done but instead $\sqrt {(x-6)^2}$ is written as $|x-6|$ and then modulus is opened for $x<6$ and $x>6$. And is said that limit not exists at $x=6$.
So in that case $\sqrt{x}$ can also be written as $|x|$. And we know that $|x|$ limit exists at $x\to 0$ then it must also exist for $\sqrt{x}$. I know that graph of $\sqrt{x}$ is not same as $|x|$. But if this method can be applied to the previous problem then why can't for this. Please explain me I am getting much confused in such type of problems.
The issue with your first one is that you have defined $h>0$, so $\sqrt{0-h}$ is in fact STILL the square root of a negative number, so the limit is undefined going from the negative side.
The reason on the second one: Consider going from the left side ($x<0$). We have $\sqrt{(x-6)^2}=6-x=|x-6|.$ This is simply because the final result of a square root must be positive if it's defined (it's obviously defined since the inside of the square root is always nonnegative). $x-6$ is not positive for $x<6$ so we must use $6-x$ in this case, or $|x-6|$ overall.