Evaluate limit of lim x → 0 ( 1 + x ) ( 1 / 5 ) − ( 1 − x ) ( 1 / 5 ) x

Question

Evaluate limit of        lim          x      →      0                  (      1      +      x              )                  (          1                      /                    5          )                    −      (      1      −      x              )                  (          1                      /                    5          )                      x

pyphekam · Accepted Answer

Start by adding and subtracting 1 in the numerator:      lim          x      →      0                  (      1      +      x              )                              1            5                              −      1      +      1      −      (      1      −      x              )                              1            5                                x  Now split up the limit like this:      lim          x      →      0                  (      1      +      x              )                              1            5                              −      1        x    +      lim          x      →      0                  (      1      −      x              )                              1            5                              −      1              −      x      Using the substitution   u  =  −  x on the second limit, you should see that both limits are the definition of the derivative of   f  (  x  )  =      x                  1        5             at   x  =  1      f    ′    (      x    0    )  =      lim          h      →      0                  f      (              x        0            +      h      )      −      f      (              x        0            )        h        f    ′    (  1  )  =      lim          h      →      0                  (      1      +      h              )                              1            5                              −              1                              1            5                                h  Using the power rule, we know that       f    ′    (  x  )  =      1    5        x          −              4        5              . So, the answer is   2      f    ′    (  1  )  =  2  (      1    5    (  1      )          −              4        5              )  =                                          2            5                    .                    Alternatively, you can see this by using the alternative definition       f    ′    (  1  )  =      lim          x      →      0                  f      (      1      +      x      )      −      f      (      1      −      x      )              2      x       and multiplying the numerator and denominator of our original limit by 2

Evaluate limit of <munder> <mo movablelimits="true" form="prefix">lim <mrow class="MJX-T

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