Question

If the normal of $y=f\left(x\right)$ at $(0,0)$ is given by $y-x=0,$ then
$\underset{x\to 0}{lim}\frac{x^2}{f\left(x^2\right)-20f\left(9x^2\right)+2f\left(99x^2\right)}$

• A. equals $\frac{1}{19}$
• B. equals $\frac{-1}{19}$
• C. equals $\frac{1}{2}$
• D. does not exist

Answer 1

The correct option is B equals $\frac{-1}{19}$

Equation of normal is $y=x$
Slope of normal at $(0,0)$ is $1$
Slope of tangent $\times$ Slope of normal $=-1$
$\Rightarrow$ Slope of tangent at $(0,0)$ is $f^{\prime }\left(0\right)=-1$

Now, $L=\underset{x\to 0}{lim}\frac{x^2}{f\left(x^2\right)-20f\left(9x^2\right)+2f\left(99x^2\right)}\left(\frac{0}{0}\text{form}\right)$
Using L'Hospital rule
$L=\underset{x\to 0}{lim}\frac{2x}{2x{f}^{\prime }\left(x^2\right)-360x{f}^{\prime }\left(9x^2\right)+396x{f}^{\prime }\left(99x^2\right)}$
$=\underset{x\to 0}{lim}\frac{1}{f^{\prime }\left(x^2\right)-180f^{\prime }\left(9x^2\right)+198f^{\prime }\left(99x^2\right)}$
$=\frac{1}{f^{\prime }\left(0\right)-180f^{\prime }\left(0\right)+198f^{\prime }\left(0\right)}$
$=\frac{1}{(1-180+198)f^{\prime }\left(0\right)}$
$=\frac{-1}{19}$