Recent content by cdaw784

  1. C

    Limit of (√6(a+h) - √6a) / h as h approaches 0 in terms of constant a

    i ended up doing, 6h / h (2√6a) + (√6h) and than cancel out the h = 6 / (2√6a) + (√6h) and then multiplying (2√6a) + (√6h) to top and bottom to get rid of sqrts = 6 [(2√6a) + (√6h)] / 12a+6h = 12(√6a)+6(√6h) / 12a + 6h if i plug in 0 for h i get, 12(√6a)+6(√6(0)) / 12a + 6(0)...
  2. C

    Limit of (√6(a+h) - √6a) / h as h approaches 0 in terms of constant a

    so, (6a+6h-6a)= 0a+6h= 6h and [h(√6(a+h) + √6a)]= (h√6(a+h)) + (h√6a) which would mean, 6h / [(h√6(a+h)) + (h√6a)]Should I separate (h√6(a+h)) into [h√6a + h√6h] or no?
  3. C

    Limit of (√6(a+h) - √6a) / h as h approaches 0 in terms of constant a

    well, I've multiplied both the top and bottom by the conjugate which would then equal, 6(a+h)-6a / h(√6(a+h) + √6a) which i then get, 6a+6h-6a/ h(√6(a+h) + √6a) but then the whole thing just equals 0 and I'm lost. am i correct or wrong so far?
  4. C

    Limit of (√6(a+h) - √6a) / h as h approaches 0 in terms of constant a

    Homework Statement Find in terms of constant a lim (√6(a+h) - √6a) / h h→0 The Attempt at a Solution I've tried multiplying by the conjugate [√6(a+h) + √6a] but I still can't seem to get the right answer. Any help would be appreciated.