Question 1117348
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Introduce  z = {{{e^x}}}.  Then the given fraction takes the form

    fraction = {{{(z^17 - 17z + 16)/(z^16 - 16z + 15)}}}.    (1)



The polynomial  {{{z^17 - 17z + 16}}}  has the root z=1  and therefore is divided by (z-1) without a remainder.

The factoring formula is  

{{{z^17 - 17z + 16}}} = {{{(z^16 + z^15 + z^14 + ellipsis + z^2 + z - 16)*(z-1)}}}.     (2)



Similarly, the polynomial  {{{z^16 - 16z + 15}}}  has the root z=1  and therefore is divided by (z-1) without a remainder.

The factoring formula is  

{{{z^16 - 16z + 15}}} = {{{(z^15 + z^14 + z^13 + ellipsis + z^2 + z - 15)*(z-1)}}}.     (3)



If you substitute (2) and (3)  into (1), you will get after canceling (z-1)

    fraction = {{{(z^16 + z^15 + z^14 + ellipsis + z^2 + z - 16)/(z^15 + z^14 + z^13 + ellipsis + z^2 + z - 15)}}}    (4)


It is still not a safe situation, since both polynomials in numerator and denominator of (4) have z= 1 as a root.


So, we need divide each of (2) and (3) by (z-1) one more time. If you do it, you will get

{{{z^16 + z^15 + z^14 + ellipsis + z^2 + z - 16}}} = {{{(z^15 + 2z^14 + 3z^13 + 4z^12 + ellipsis + 15z + 16)*(z-1)}}},   (5)

{{{z^15 + z^14 + z^13 + ellipsis + z^2 + z - 15}}} = {{{(z^14 + 2z^13 + 3z^12 + 4z^11 + ellipsis + 14z + 15)*(z-1)}}}.   (6)



Hence, when you substitute (5) and (6) into (4) and cancel the common factor (z-1) again, you will get

    fraction = {{{(z^15 + 2z^14 + 3z^13 + 4z^12 + ellipsis + 15z + 16)/(z^14 + 2z^13 + 3z^12 + 4z^11 + ellipsis + 14z + 15)}}}   (7)


Now you can safely find the limit of (7) at z ---> 1  simply substituting z = 1 into its numerator and denominator. You will get

    fraction limit at z --> 1 is equal to  = {{{(1 + 2 + 3 + 4 + ellipsis + 15 + 16)/(1 + 2 + 3 + 4 + ellipsis + 14 + 15))}}}   (8)


Easy summation of arithmetic progressions gives  Numerator = {{{(16*17)/2}}} = 136,  Denominator = {{{(15*16)/2}}} = 120.


Hence the answer is:  The given fraction limit at x ---> 0 is   {{{136/120}}} = {{{17/15}}}.
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