Question

The graph of the function $y=f\left(x\right)$ has unique tangent at the point $(a,0)$ through which the graph passes. Then $\underset{x\to a}{lim}\frac{\ln (1+6f(x\left)\right)}{3f\left(x\right)}$ is equal to

Answer 1

Given $f\left(x\right)$ has unique tangent at $(a,0)$
$\therefore f\left(a\right)=0$ and $f\text{'}\left(x\right)$ exists.

$\underset{x\to a}{lim}\frac{\ln (1+6f(x\left)\right)}{3f\left(x\right)}\left(\frac{0}{0}\text{form}\right)$
Applying L Hospital's rule
$=\underset{x\to a}{lim}\frac{\frac{6f^{\prime }\left(x\right)}{1+6f\left(x\right)}}{3f^{\prime }\left(x\right)}$
$=\frac{2}{1+6f\left(x\right)}[\because f(a)=0]=2$