We have,
x→0lim(2−cosxcos2x)(x2x+2)→1∞
Hence,
=ex→0lim(x21−cosxcos2x)×(x+2)
Now,
x→0lim(x21−cosxcos2x)→(00)
Using L'Hospital Rule
x→0lim2xsinxcos2x−cosx×2cos2x1×(−2sin2x)
=x→0lim2xcos2xsinxcos2x+sin2x⋅cosx
=x→0lim2xcos2xsin3x
=23x→0lim3xsin3x×x→0limcos2x1
=23x→0lim3xsin3x×1
=23
So,
ex→0lim(x21−cosxcos2x)(x+2)
=e23×2=e3
⇒a=3