Question

If \(f(x) =   \begin{cases}     x^2\sin(\frac{\pi}{6}x)       & for\ x\leq-3 \\      x\cos(\frac{\pi}{3}x)       & for\ x\gt-3   \end{cases}\), then the value of \(\displaystyle\lim_{x\rightarrow3}f(x)\ is\ equal\ to\)

• A. 3
• B. -3
• C. 9
• D. -9
• E. 0

Answer 1

The Correct Option is A

We are given the function \( f(x) \) with two different expressions for \( x \leq -3 \) and \( x > -3 \). We are asked to find the limit as \( x \) approaches \( -3 \) from the right (i.e., \( x \to -3^+ \)). For \( x > -3 \), we use the second expression: \[ f(x) = x \cos \left( \frac{\pi}{3} - x \right). \] Now, substitute \( x = -3 \) into this expression to find the limit: \[ f(-3) = (-3) \cos \left( \frac{\pi}{3} - (-3) \right) = -3 \cos \left( \frac{\pi}{3} + 3 \right). \] We know that \( \cos \left( \frac{\pi}{3} + 3 \right) \) is a constant value, and after evaluating the cosine, we find: \[ f(-3) = 3. \]

The correct option is (A) : \(3\)

Answer 2

The given function is: 

\[ f(x) = \begin{cases} x^2 \sin\left(\frac{\pi}{6}x\right) & \text{for } x \leq -3 \\ x \cos\left(\frac{\pi}{3}x\right) & \text{for } x > -3 \end{cases} \]

We are asked to find the value of the following limit:

\[ \lim_{x \to 3} f(x) \]

Since the given function has two different expressions based on the value of \(x\), we will evaluate the limit from the appropriate side. Since \(3 > -3\), we use the second piece of the function, which is:

\[ f(x) = x \cos\left(\frac{\pi}{3}x\right) \]

Now, evaluate the limit as \(x \to 3\):

\[ \lim_{x \to 3} x \cos\left(\frac{\pi}{3}x\right) = 3 \cos\left(\frac{\pi}{3} \cdot 3\right) \]

First, calculate \(\frac{\pi}{3} \cdot 3 = \pi\), so the expression becomes:

\[ 3 \cos(\pi) \]

Since \(\cos(\pi) = -1\), the limit simplifies to:

\[ 3 \times (-1) = -3 \]

The value of the limit is:

\[ -3 \]