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 \( f(x) = \int_0^x t e^{2t} \sqrt{1 + 3t^2} \, dt \).

### Step 1: Differentiate \( g(x) \)

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

### Step 2: Differentiate \( f(x) \)

Using the Fundamental Theorem of Calculus:
\[ f'(x) = x e^{2x} \sqrt{1 + 3x^2} \]

### Step 3: Set Up the Limit

Now we can substitute \( f'(x) \) and \( g'(x) \) into the limit:
\[ L = \lim_{x \to \infty} \frac{x e^{2x} \sqrt{1 + 3x^2}}{e^{Cx} (c x^{c-1} + C x^c)} \]

### Step 4: Simplify the Limit

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

### Step 5: Analyze the Exponential Term

For \( L \) to be a non-zero finite number, the exponent \( 2 - C \) must equal zero:
\[ 2 - C = 0 \implies C = 2 \]

### Step 6: Substitute \( C = 2 \) into the Limit

Substituting \( C = 2 \):
\[ L = \lim_{x \to \infty} \frac{x \sqrt{1 + 3x^2}}{c x^{c-1} + 2 x^c} \]
As \( x \to \infty \), the dominant term in the denominator is \( 2x^c \):
\[ L = \lim_{x \to \infty} \frac{x \sqrt{1 + 3x^2}}{2 x^c} \]
This simplifies to:
\[ L = \lim_{x \to \infty} \frac{\sqrt{1 + 3x^2}}{2 x^{c-1}} = \frac{\sqrt{3}}{2} \text{ if } c = 2 \]

### Final Result

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