Sure, let's go through the limits step-by-step and determine their values.
### 1. [tex]\( \lim_{h \to 0} \left( \frac{\sqrt{x h} - \sqrt{x}}{h} \right) \)[/tex]
This is the limit as [tex]\( h \)[/tex] approaches 0. When we substitute [tex]\( h = 0 \)[/tex] in the expression, we get:
[tex]\[ \frac{\sqrt{x \cdot 0} - \sqrt{x}}{0} = \frac{0 - \sqrt{x}}{0} = \frac{-\sqrt{x}}{0} = \text{undefined (nan)} \][/tex]
So, the limit is:
[tex]\[ \boxed{\text{nan}} \][/tex]
### 2. [tex]\( \lim_{x \to a} \frac{\sqrt{x a} - \sqrt{3x - a}}{x - a} \)[/tex]
When [tex]\( x \)[/tex] approaches [tex]\( a \)[/tex], the expression in the limit becomes:
[tex]\[ \frac{\sqrt{a^2} - \sqrt{3a - a}}{a - a} = \frac{|a| - \sqrt{2a}}{0} = \text{undefined (can lead to } -\infty \text{)} \][/tex]
Hence, the limit is:
[tex]\[ \boxed{-\infty} \][/tex]
### 3. [tex]\( \lim_{x \to 1} \frac{\sqrt{2x} - \sqrt{3 - x^2}}{x - 1} \)[/tex]
When [tex]\( x \)[/tex] approaches 1, the expression [tex]\( \sqrt{2x} - \sqrt{3 - x^2} \)[/tex] needs to be evaluated. This is a case where the numerator results in an indeterminate form and needs to be evaluated carefully.
Substituting [tex]\( x = 1 \)[/tex]:
[tex]\[ \frac{\sqrt{2 \cdot 1} - \sqrt{3 - 1^2}}{1 - 1} = \frac{\sqrt{2} - \sqrt{2}}{0} = \frac{0}{0} \][/tex]
By analyzing more deeply (L'Hôpital's Rule may be applicable for evaluating such limits), we can deduce that the limit evaluates to:
[tex]\[ \boxed{\sqrt{2}} \][/tex]
### 4. [tex]\( \lim_{x \to 1} \frac{x - 1}{\sqrt{x^2 \cdot 3} - 2} \)[/tex]
As [tex]\( x \)[/tex] approaches 1, the limit is:
[tex]\[ \frac{1 - 1}{\sqrt{1^2 \cdot 3} - 2} = \frac{0}{\sqrt{3} - 2} = 0 \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{0} \][/tex]
### 5. [tex]\( \lim_{x \to 0} \frac{\sqrt{x^2} - \sqrt{1 - x^2}}{\frac{2x^2}{2}} \)[/tex]
When [tex]\( x \)[/tex] approaches 0:
[tex]\[ \frac{\sqrt{0^2} - \sqrt{1 - 0^2}}{\frac{2 \cdot 0^2}{2}} = \frac{0 - 1}{0} = \frac{-1}{0} = -\infty \][/tex]
Thus, the limit is:
[tex]\[ \boxed{-\infty} \][/tex]
### 6. [tex]\( \lim_{x \to 1} \frac{x - \sqrt{2 - x^2}}{2x - \sqrt{2x^2}} \)[/tex]
Substituting [tex]\( x = 1 \)[/tex] into the expression:
[tex]\[ \frac{1 - \sqrt{2 - 1^2}}{2 \cdot 1 - \sqrt{2 \cdot 1^2}} = \frac{1 - \sqrt{1}}{2 - \sqrt{2}} = \frac{1 - 1}{2 - \sqrt{2}} = \frac{0}{2 - \sqrt{2}} = 0 \][/tex]
Therefore, the limit is:
[tex]\[ \boxed{0} \][/tex]
