Two short calculus questions - domain and limits

Part 1.

For the formula to make sense, we need the following:

• $3 - x^2 > 0$
• $2x-5 \neq 0$
• $x > 0$
• $\log x \neq 0$

These conditions translate to

• $ -\sqrt{3} < x < \sqrt{3}$
• $ x \neq \frac{5}{2}$
• $x > 0$
• $x \neq 1$

so, $D = (0, 1) \cup (1, \sqrt{3})$.

Part 2.

Note that $x^2 - 5x + 4 = (x-1)(x-4)$. For $x \rightarrow 1$, the left limit is $-3\sqrt{2-1} = -3\sqrt{1} = -3$, and the right limit is $1-4 = -3$, as we can simplify $(x-1)$. Since they are equal, the limit exists, and it is equal to $-3$. Since $f(1) = 7 \neq -3$, $f$ is not continuous at $1$.