limx→0f(x)x=10
So, f(0) must be 0.
limx→0f(x)x (Indeterminate form)
Using L' Hospitals' rule, we get
limx→0f′(x)1=10
⇒ f′(0)=10 ...(1)
Now, limx→0f(sin x)g(x) cos2 x→Indeterminate form
Using L' Hospitals' rule, we get
limx→0f′(sin x)×cos xg′(x)cos2 x+g(x)×2 cos x×(−sin x)=5
⇒f′(0)g′(0)=5⇒g′(0)=105=2 ...(2)
limx→0g(1−cos x)x2→ Indeterminate form
=limx→0g′(1−cos x)×sin x2x=g′(0)2=22=1