To determine

To evaluate: limx→0ex−e−x−2xx−sinx

Answer

limx→0ex−e−x−2xx−sinx=2

Explanation

Calculations:

limx→0ex−e−x−2xx−sinx (00Form)Applying l'Hospital's rule, we get=limx→0ex+e−x−21−cosx (00Form)Applying l'Hospital's rule again, we get=limx→0ex−e−xsinx=limx→0ex+e−xcosx=e0+e−0cos(0)=1+11=2

Conclusion: The value of the limit is limx→0ex−e−x−2xx−sinx=2