I am trying to solve this problem : P = lim x → 0 p H + ( 1 − p H ) ( 1 − x ) t p H + ( 1 − p H ) ( 1 − x ) t − x − p L + ( 1 − p L ) ( 1 − x ) t p L + ( 1 − p L ) ( 1 − x ) t − x x

Question

I am trying to solve this problem :      P    =      lim          x      →      0                                                p                          H                                +          (          1          −                      p                          H                                )          (          1          −          x                      )                          t                                                            p                          H                                +          (          1          −                      p                          H                                )          (          1          −          x                      )                          t              −              x                                          −                                    p                          L                                +          (          1          −                      p                          L                                )          (          1          −          x                      )                          t                                                            p                          L                                +          (          1          −                      p                          L                                )          (          1          −          x                      )                          t              −              x                                            x

iskakanjulc · Accepted Answer

Let   a  =      p    H   and   b  =      p    L  ; so the numerator of the expression is  A  =            a      +      (      1      −      a      )      (      1      −      x              )        t                    a      +      (      1      −      a      )      (      1      −      x              )                  t          −          x                      −            b      +      (      1      −      b      )      (      1      −      x              )        t                    b      +      (      1      −      b      )      (      1      −      x              )                  t          −          x                    Using Taylor expansion around   x  =  0  (  1  −  x      )    k    =      ∑          n      =      0        ∞    (  −  1      )    n                      (                    k      n                      )                  x    n  Use it for a very few terms and continue with the long division to obtain            a      +      (      1      −      a      )      (      1      −      x              )        t                    a      +      (      1      −      a      )      (      1      −      x              )                  t          −          x                      =  1  +  (  a  −  1  )      x    2    −      1    2    (  a  −  1  )  (  2  a  t  −  1  )      x    3    +  O      (          x      4        )  that is to say  A  =  (  a  −  b  )      x    2    +  O      (          x      3        )      ⟹        P    =      lim          x      →      0            A    x    =      lim          x      →      0        [  (  a  −  b  )  x  ]  =  0that is to say  A  =  (  a  −  b  )      x    2    +  O      (          x      3        )      ⟹        P    =      lim          x      →      0            A    x    =      lim          x      →      0        [  (  a  −  b  )  x  ]  =  0which is what you found using L&#039;Hospital.

I am trying to solve this problem : <mrow class="MJX-TeXAtom-ORD"> <mi mathvariant="double-s

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New Questions in Calculus 1