limn→∞(1+sinan)n
Take 1n=x
As n→∞,x→0
⇒limx→0(1+sinax)1x
This is in the form of 1∞
We know that limx→afg=elimx→(f−1)g
Here f=1+sinax and g=1x
⇒elimx→0(1+sinax−1)1x
⇒elimx→0(sinax)aax
⇒ea (sincelimx→0sinxx=1)