Question

If $f^{\prime }\left(0\right)=3,\underset{x\to 0}{lim}\frac{x^2}{f\left(x^2\right)-6f\left(4x^2\right)+5f\left(7x^2\right)}=$

• A. $\frac{1}{36}$
• B. $-\frac{1}{36}$
• C. $\frac{1}{34}$
• D. $\frac{1}{106}$

Answer 1

The correct option is A $\frac{1}{36}$

Since, the function is of $\frac{0}{0}$ form on applying the limits,
We apply L-Hospital's Rule and differentiate the numerator and denominator individually,

$\underset{x\to 0}{lim}\frac{2x}{2x{f}^{\prime }\left(x^2\right)-48x{f}^{\prime }\left(x^2\right)+70x{f}^{\prime }\left(x^2\right)}$
Applying the limit, we again get the $\frac{0}{0}$ form
Again using L-Hospital's Rule,

$\underset{x\to 0}{lim}\frac{2}{2f^{\prime }\left(x^2\right)+4x^2{f}^{\prime \prime }\left(x^2\right)-48f^{\prime }\left(x^2\right)-2304x^2{f}^{\prime \prime }\left(x^2\right)+70f^{\prime }\left(x^2\right)+4900x^2{f}^{\prime \prime }\left(x^2\right)}$

$=\frac{2}{6-144+210}=\frac{2}{72}=\frac{1}{36}$