Find `lim_(x to 0)f(x)` and `lim_(x to 1)f(x)`, where
`f(x)={{:(2x+3",", x le 0), (3(x+1)",", x gt 0):}`

Answer

To find the limits of the function \( f(x) \) as \( x \) approaches 0 and 1, we will evaluate the piecewise function defined as follows:

\[ f(x) = \begin{cases} 2x + 3 & \text{if } x \leq 0 \\ 3(x + 1) & \text{if } x > 0 \end{cases} \]

### Step 1: Find \( \lim_{x \to 0} f(x) \)

Since we are approaching \( x = 0 \), we need to consider the value of the function from both sides of 0.

- From the left (as \( x \) approaches 0 from negative values), we use the first piece of the function:
\[ f(x) = 2x + 3 \quad \text{for } x \leq 0 \]
Calculating the limit:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0} (2x + 3) = 2(0) + 3 = 3 \]

- From the right (as \( x \) approaches 0 from positive values), we use the second piece of the function:
\[ f(x) = 3(x + 1) \quad \text{for } x > 0 \]
Calculating the limit:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0} (3(x + 1)) = 3(0 + 1) = 3 \]

Since both one-sided limits are equal, we conclude:
\[ \lim_{x \to 0} f(x) = 3 \]

### Step 2: Find \( \lim_{x \to 1} f(x) \)

Now we need to find the limit as \( x \) approaches 1. Since \( 1 > 0 \), we will use the second piece of the function:
\[ f(x) = 3(x + 1) \quad \text{for } x > 0 \]
Calculating the limit:
\[ \lim_{x \to 1} f(x) = 3(1 + 1) = 3 \cdot 2 = 6 \]

### Final Results
Thus, we have:
\[ \lim_{x \to 0} f(x) = 3 \]
\[ \lim_{x \to 1} f(x) = 6 \]