We have limt→x+1t2f(x+1)−(x+1)2f(t)f(t)−f(x+1)=1.
Now, applying L'Hospital rule, (Differentiate Numerator and Denominator w.r.t. t).
limt→x+12tf(x+1)−(x+1)2f′(t)f′(t)=1
or,2(x+1)f(x+1)−(x+1)2f′(x+1)f′(x+1)=1
or, 2(x+1)f(x+1)−(x+1)2f′(x+1)=f′(x+1)
Which is equivalent to
2xf(x)−x2f′(x)=f′(x)
or, (1+x2)f′(x)=2xf(x)
or, f′(x)f(x)=2x1+x2
Integrating we have,
f(x)=c(1+x2)
Given f(0)=1
⇒c=1
∴f(x)=(1+x2).
Now,
limx→1ln(f(x))−ln2x−1
=limx→1ln(1+x2)−ln2x−1
Again using L'Hospital rule ,
=limx→12x1+x2=1