It is given that f'(x) = 3x^2 and g'(x) = 2x where f'(x) and g'(x) are first order derivatives of f(x) and g(x) . What is the value of x 1 f(x)g(x) by using L Hospital's rule. a) 32 b) 13 c) 12 d) 16

To find the value of ${\lim }_{x\to 1}\frac{f(x)}{g(x)}$ using L'Hôpital's rule, we proceed as follows:

Given: $$f^{\prime }(x)=3x^2,g^{\prime }(x)=2x$$

First, check if the limit is of the indeterminate form $\frac{0}{0}$ or $\frac{\infty }{\infty }$ at $x=1$: $$f(1)=\int f^{\prime }(x)dx=\int 3x^{2}dx=x^3+C_1⟹f(1)=1^3+C_1=1+C_1$$ $$g(1)=\int g^{\prime }(x)dx=\int 2xdx=x^2+C_2⟹g(1)=1^2+C_2=1+C_2$$

Since $f(1)$ and $g(1)$ are finite, we apply L'Hôpital's rule: $$\underset{x\to 1}{\lim }\frac{f(x)}{g(x)}=\underset{x\to 1}{\lim }\frac{f^{\prime }(x)}{g^{\prime }(x)}=\frac{f^{\prime }(1)}{g^{\prime }(1)}=\frac{3(1{)}^2}{2(1)}=\frac{3}{2}$$

Answer: $\boxed{\frac{3}{2}}$ (Option a)