If `f(x)={("sin"(pix)/2,, x lt1),(4x-3,,1lexle2),(log2(2x^(2)-4),,2ltxlt3):}` then value of `lim_(xto1)f(x)+lim_(xto2^(+))f(x)` is

Options

A. 2
B. 3
C. 4
D. 5

Answer

To solve the problem, we need to evaluate the limits of the piecewise function \( f(x) \) at specific points and then sum those limits. The function is defined as follows:

\[ f(x) = \begin{cases} \frac{\sin(\pi x)}{2} & \text{if } x < 1 \\ 4x - 3 & \text{if } 1 \leq x \leq 2 \\ \log_2(2x^2 - 4) & \text{if } 2 < x < 3 \end{cases} \]

We need to find the value of \( \lim_{x \to 1} f(x) + \lim_{x \to 2^+} f(x) \).

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

Since \( x \) approaches 1 from the left, we use the first case of the piecewise function:

\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{\sin(\pi x)}{2} \]

Substituting \( x = 1 \):

\[ \lim_{x \to 1} f(x) = \frac{\sin(\pi \cdot 1)}{2} = \frac{\sin(\pi)}{2} = \frac{0}{2} = 0 \]

### Step 2: Evaluate \( \lim_{x \to 2^+} f(x) \)

Since \( x \) approaches 2 from the right, we use the third case of the piecewise function:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} \log_2(2x^2 - 4) \]

Substituting \( x = 2 \):

\[ \lim_{x \to 2^+} f(x) = \log_2(2 \cdot 2^2 - 4) = \log_2(2 \cdot 4 - 4) = \log_2(8 - 4) = \log_2(4) \]

We know that \( \log_2(4) = 2 \) because \( 4 = 2^2 \).

### Step 3: Combine the limits

Now we can sum the two limits:

\[ \lim_{x \to 1} f(x) + \lim_{x \to 2^+} f(x) = 0 + 2 = 2 \]

### Final Answer

The value of \( \lim_{x \to 1} f(x) + \lim_{x \to 2^+} f(x) \) is \( 2 \).