Question

If \(\lim_{x \to 3} \frac{(2x - k)\tan(x - 3)}{x^2 - 6x + 9} = 2\), then the value of 'k' is equal to 
 

• A. 2
• B. 3
• C. 4
• D. 6
• E. 8

Hint:

For a limit \(\lim_{x \to c} \frac{f(x)}{g(x)}\) to exist and be finite when \(g(c) = 0\), \(f(c)\) must also be 0. Setting the numerator to zero at the limit point is a common trick to find unknown constants.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We are given a limit involving a trigonometric function and a quadratic expression.
The denominator \(x^2 - 6x + 9\) can be rewritten as \((x-3)^2\).
We know that \(\lim_{\theta \to 0} \frac{\tan \theta}{\theta} = 1\).

Step 2: Key Formula or Approach:
As \(x \to 3\), let \(u = x - 3\). Then \(u \to 0\).
Substitute \(x = u + 3\) into the limit expression.

Step 3: Detailed Explanation:
Rewrite the denominator and separate the limit:
\[ \lim_{x \to 3} \frac{(2x - k) \tan(x - 3)}{(x - 3)^2} = 2 \]
\[ \lim_{x \to 3} \left[ \frac{\tan(x - 3)}{x - 3} \cdot \frac{2x - k}{x - 3} \right] = 2 \]
We know \(\lim_{x \to 3} \frac{\tan(x - 3)}{x - 3} = 1\).
So the equation simplifies to:
\[ \lim_{x \to 3} \frac{2x - k}{x - 3} = 2 \]
For this limit to be a finite non-zero value, the numerator must be zero at \(x = 3\) (0/0 indeterminate form).
\[ 2(3) - k = 0 \implies 6 - k = 0 \implies k = 6 \]
Let's verify: if \(k = 6\), the expression is \(\frac{2(x-3)}{x-3} = 2\), which matches the given limit value.

Step 4: Final Answer:
The value of \(k\) is 6.