Question

Let $g\left(x\right)=\left|\begin{array}{ccc}f(x+c)& f(x+2c)& f(x+3c)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|$, where $c$ is constant then $\underset{x\to 0}{lim}\frac{g\left(x\right)}{x}$ is equal to

• A. $0$
• B. $1$
• C. $-1$
• D. $f\left(c\right)$

Answer 1

The correct option is A $0$

Using L-Hospital rule,
$\underset{x\to 0}{lim}\frac{g\left(x\right)}{x}=\underset{x\to 0}{lim}\frac{g^{\prime }\left(x\right)}{1}=g^{\prime }\left(0\right)$
Now
$g^{\prime }\left(x\right)=\left|\begin{array}{ccc}{f}^{\prime }(x+c)& f^{\prime }(x+2c)& f^{\prime }(x+3c)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|$
$\Rightarrow g^{\prime }\left(0\right)=\left|\begin{array}{ccc}{f}^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\\ f\left(c\right)& f\left(2c\right)& f\left(3c\right)\\ f^{\prime }\left(c\right)& f^{\prime }\left(2c\right)& f^{\prime }\left(3c\right)\end{array}\right|=0$, Since $R_1=R_3$
Hence value of required limit is $0$.