Evaluate lim x tends to y { (e^sinx - e^siny)/(x-y) }?

Answer 1

Evaluate  lim x tends to y { (e^sinx - e^siny)/(x-y) }?Put x= y +dy. dy is very term. When x tend to y, dy tends to zero.

Put it now e^(sin(y+dy) - e^ sin y)

expand sin y +dy= sin y cos dy+ cos y sin dy. When dy tend to 0, we have sin y+dy= sin y + cos y dy.

The term e^sinx - e^siny= e^ sin y e^cos y dy-e^sin y= e^sin y( e^cosy dy -1)= e^ siny ( (1+cos y dy-1)

e^ sin y cos y dy. Divide this by (x-y)=dy. The answer you get it  cos y e^sin y. This is the answer.

Answer 2

Hint: Saying x --> y can be said as y = x + ε with ε -->0. 

Work at it a little now.

Conclusions? Got it? if not, 

Show your work if need be and we can continue. 

Answer 3

It's easy if you recognize that as the derivative f'(y) of where f(t) = e^(sin t). The chain rule gives you:

   f'(t)  =  d/dt e^(sin t)  =  e^(sin t) * d/dt (sin t) 

   f'(t)  =  e^(sin t) * cos t

Then f'(y) = e^(sin y) * cos y.

Solving this from first principles seems like it would be much messier.