t→xlimt−xt2f2(x)−x2f2(t)=0
Using L'Hospital
t→xlim12tf2(x)−x22f(t)f′(t)=0
⇒2xf2(x)−x22f(x)f′(x)=0
⇒2xf(x)[f(x)−xf′(x)]=0
But f(x)=0
So, xf′(x)=f(x)
⇒xdxdy=y
⇒y1dy=x1dx
Integration gives
lny=lnx+lnc
⇒y=cx⇒f(x)=cx
Now
f(1)=c=e (given)
So, f(x)=ex
Now if f(x)=1, then ex=1⇒x=e1