When dealing with \([x]\) in limits, always think about the specific interval just after or just before the limit point. Here, for \(x \to 2^+\), \(x\) is slightly more than 2, so \([x]\) is exactly 2.
Step 1: Understanding the Concept:
For a function to be continuous at \(x = 2\), the left-hand limit (LHL), the right-hand limit (RHL), and the value of the function at \(x = 2\) must all be equal. Step 3: Detailed Explanation:
1. Left Hand Limit (LHL) at \(x = 2\):
For \(x<2\), \(x^2 - 5x + 6 = (x-2)(x-3)\).
As \(x \to 2^-\), \((x-2)\) is negative and \((x-3)\) is negative, so their product is positive.
Thus \(|x^2 - 5x + 6| = x^2 - 5x + 6\).
LHL $= \lim_{x \to 2^-} \frac{\lambda (x^2 - 5x + 6)}{-\mu (x^2 - 5x + 6)} = -\frac{\lambda}{\mu}$.
2. Right Hand Limit (RHL) at \(x = 2\):
For \(x \to 2^+\), \(x \in (2, 3)\), so \([x] = 2\).
RHL $= \lim_{x \to 2^+} e^{\frac{\tan(x-2)}{x-2}}$.
Using the standard limit \(\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1\), the exponent becomes 1.
RHL $= e^1 = e$.
3. Value of function at \(x = 2\):
\(f(2) = \mu\).
For continuity: \(LHL = RHL = f(2)\).
\(\mu = e\) and \(-\frac{\lambda}{\mu} = e \Rightarrow \lambda = -e^2\).
Sum \(\lambda + \mu = -e^2 + e = e(-e + 1)\). Step 4: Final Answer:
The value is \(e(-e+1)\).
