Let `g (x ) = x ^(C )e ^(Cx) and f (x) = int _(0)^(x) te ^(2t) (1+3t ^(2))^(1//2) dt.` If `L = lim _(x to oo) (f'(x ))/(g '(x))` is non-zero finite number then :
The value of L . Is :

Options

A. `2/7`
B. `1/2`
C. `(sqrt3)/(4)`
D. `(sqrt3)/(2)`

Answer

To solve the problem, we need to find the limit \( L = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} \) where \( g(x) = x^c e^{Cx} \) and \( f(x) = \int_0^x t e^{2t} (1 + 3t^2)^{1/2} dt \).

### Step 1: Find \( g'(x) \)

Given:
\[ g(x) = x^c e^{Cx} \]

Using the product rule:
\[ g'(x) = \frac{d}{dx}(x^c) \cdot e^{Cx} + x^c \cdot \frac{d}{dx}(e^{Cx}) \]
\[ = c x^{c-1} e^{Cx} + x^c \cdot C e^{Cx} \]
\[ = e^{Cx} \left( c x^{c-1} + C x^c \right) \]
\[ = e^{Cx} x^{c-1} (c + Cx) \]

### Step 2: Find \( f'(x) \)

Using Leibniz's rule for differentiation under the integral sign:
\[ f'(x) = \frac{d}{dx} \int_0^x t e^{2t} (1 + 3t^2)^{1/2} dt = x e^{2x} (1 + 3x^2)^{1/2} \]

### Step 3: Set up the limit \( L \)

Now we can express \( L \):
\[ L = \lim_{x \to \infty} \frac{f'(x)}{g'(x)} = \lim_{x \to \infty} \frac{x e^{2x} (1 + 3x^2)^{1/2}}{e^{Cx} x^{c-1} (c + Cx)} \]

### Step 4: Simplify the limit

We can simplify this limit:
\[ L = \lim_{x \to \infty} \frac{x e^{2x} (1 + 3x^2)^{1/2}}{e^{Cx} x^{c-1} (c + Cx)} = \lim_{x \to \infty} \frac{(1 + 3x^2)^{1/2}}{x^{c-2} (c + Cx) e^{(C-2)x}} \]

### Step 5: Analyze the limit

For \( L \) to be a non-zero finite number, we need to balance the powers of \( x \) and the exponential terms. This leads to:
\[ C - 2 = 0 \quad \Rightarrow \quad C = 2 \]
\[ c - 2 = 0 \quad \Rightarrow \quad c = 2 \]

### Step 6: Substitute \( C \) and \( c \)

Substituting \( C = 2 \) and \( c = 2 \) into the limit:
\[ L = \lim_{x \to \infty} \frac{(1 + 3x^2)^{1/2}}{x^0 (2 + 2x) e^{0}} = \lim_{x \to \infty} \frac{(1 + 3x^2)^{1/2}}{2 + 2x} \]

### Step 7: Evaluate the limit

As \( x \to \infty \):
\[ L = \lim_{x \to \infty} \frac{\sqrt{3} x}{2x} = \frac{\sqrt{3}}{2} \]

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