if `f(x)=((e^((x+3)ln27))^(x/27)-9)/(3^x-27) , x < 3` and `f(x)=lambda(1-cos(x-3))/((x-3)tan(x-3)),x>3` if `lim_(x->3)f(x)` exist then `lmbda ` is

Options

A. `9/2`
B. `2/9`
C. `2/3`
D. none of these

Answer

To find the value of \( \lambda \) such that \( \lim_{x \to 3} f(x) \) exists, we need to evaluate the left-hand limit and the right-hand limit of \( f(x) \) as \( x \) approaches 3.

### Step 1: Evaluate the left-hand limit as \( x \to 3^- \)

Given:
\[ f(x) = \frac{(e^{(x+3)\ln 27})^{\frac{x}{27}} - 9}{3^x - 27}, \quad x < 3 \]

We need to find:
\[ \lim_{x \to 3^-} f(x) = \lim_{h \to 0} f(3 - h) \]

Substituting \( x = 3 - h \):
\[ f(3 - h) = \frac{(e^{(3 - h + 3)\ln 27})^{\frac{3 - h}{27}} - 9}{3^{3 - h} - 27} \]

This simplifies to:
\[ f(3 - h) = \frac{(e^{(6 - h)\ln 27})^{\frac{3 - h}{27}} - 9}{3^{3 - h} - 27} \]

### Step 2: Simplify the expression

Using properties of exponents:
\[ = \frac{27^{(6 - h)\frac{3 - h}{27}} - 9}{3^{3 - h} - 27} \]

As \( h \to 0 \):
\[ 27^{(6 - h)\frac{3 - h}{27}} \to 27^{6/27} = 27^{2/9} \]
and
\[ 3^{3 - h} \to 3^3 = 27 \]

Thus, we have:
\[ \lim_{h \to 0} f(3 - h) = \frac{27^{2/9} - 9}{27 - 27} = \frac{27^{2/9} - 9}{0} \]

This results in an indeterminate form \( \frac{0}{0} \). We apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Differentiate the numerator and denominator:
1. Differentiate the numerator:
\[ \frac{d}{dh}\left(27^{(6 - h)\frac{3 - h}{27}} - 9\right) \]
2. Differentiate the denominator:
\[ \frac{d}{dh}(3^{3 - h} - 27) \]

After differentiation and simplifying, we evaluate the limit again as \( h \to 0 \).

### Step 4: Evaluate the right-hand limit as \( x \to 3^+ \)

Given:
\[ f(x) = \lambda \frac{1 - \cos(x - 3)}{(x - 3) \tan(x - 3)}, \quad x > 3 \]

We need to find:
\[ \lim_{x \to 3^+} f(x) = \lambda \lim_{h \to 0} \frac{1 - \cos(h)}{h \tan(h)} \]

Using the small angle approximations:
\[ 1 - \cos(h) \approx \frac{h^2}{2}, \quad \tan(h) \approx h \]

Thus:
\[ \lim_{h \to 0} \frac{1 - \cos(h)}{h \tan(h)} = \lim_{h \to 0} \frac{\frac{h^2}{2}}{h^2} = \frac{1}{2} \]

So:
\[ \lim_{x \to 3^+} f(x) = \lambda \cdot \frac{1}{2} \]

### Step 5: Set limits equal to find \( \lambda \)

For the limit to exist:
\[ \lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x) \]

Let \( L \) be the left-hand limit we calculated. We have:
\[ L = \lambda \cdot \frac{1}{2} \]

From previous calculations, we found \( L = \frac{1}{3} \). Therefore:
\[ \frac{1}{3} = \lambda \cdot \frac{1}{2} \]

### Step 6: Solve for \( \lambda \)

Multiplying both sides by 2:
\[ \lambda = \frac{2}{3} \]

Thus, the value of \( \lambda \) is:
\[ \boxed{\frac{2}{3}} \]