Question

Let $f\left(x\right)$ be differentiable function in $[-1,\infty )$ and $f\left(0\right)=1$ such that $\underset{t\to x+1}{lim}\frac{t^{2}f(x+1)-(x+1{)}^{2}f(t)}{f\left(t\right)-f(x+1)}=1$. Find the value of $\underset{x\to 1}{lim}\frac{\ln \left(f\right(x\left)\right)-\ln 2}{x-1}$

Answer 1

We have $\underset{t\to x+1}{lim}\frac{t^{2}f(x+1)-(x+1{)}^{2}f(t)}{f\left(t\right)-f(x+1)}=1$.
Now, applying L'Hospital rule, (Differentiate Numerator and Denominator w.r.t. $t$).
$\underset{t\to x+1}{lim}\frac{2tf(x+1)-(x+1{)}^2{f}^{\prime }(t)}{f^{\prime }\left(t\right)}=1$
$or,\frac{2(x+1)f(x+1)-(x+1{)}^2{f}^{\prime }(x+1)}{f^{\prime }(x+1)}=1$
$or,2(x+1)f(x+1)-(x+1{)}^2{f}^{\prime }(x+1)=f^{\prime }(x+1)$
Which is equivalent to
$2xf\left(x\right)-x^2{f}^{\prime }\left(x\right)=f^{\prime }\left(x\right)$
$or,(1+x^2)f^{\prime }\left(x\right)=2xf\left(x\right)$
$or,\frac{f^{\prime }\left(x\right)}{f\left(x\right)}=\frac{2x}{1+x^2}$
Integrating we have,
$f\left(x\right)=c(1+x^2)$
Given $f\left(0\right)=1$
$\Rightarrow c=1$
$\therefore f\left(x\right)=(1+x^2)$.
Now,
$\underset{x\to 1}{lim}\frac{\ln \left(f\right(x\left)\right)-\ln 2}{x-1}$
$=\underset{x\to 1}{lim}\frac{\ln (1+x^2)-\ln 2}{x-1}$
Again using L'Hospital rule ,
$=\underset{x\to 1}{lim}\frac{2x}{1+x^2}=1$