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

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  • Vaman
    Lv 7
    1 month ago

    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.

  • rotchm
    Lv 7
    1 month ago

    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. 

  • 1 month ago

    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.

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