# What is limit of the function (x^5+1)/(x^7+1), x-->-1 ?

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We need to find: lim x-->-1 [(x^5+1)/(x^7+1]

substituting x = -1 gives 0/0 which is indeterminate. We can use l'Hopital's rule and replace the numerator and the denominator by their derivatives.

=> lim x-->-1 [(5x^4)/(7x^6]

substitute x = -1

=> 5/7

The value of the required limit is 5/7

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

To evaluate the limit, we'll re-write the numerator and denominator, using the identity:

a^n + b^n = (a+b)*(a^(n-1) - a^(n-2)*b + a^(n-3)*b^2 - a^(n-4)*b^3 + ....)

According to this formula, we'll get:

x^5 + 1 = (x + 1)(x^4 - x^3 + x^2 - x + 1)

x^7 + 1 = (x + 1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)

(x^5+1)/(x^7+1) = (x^4 - x^3 + x^2 - x + 1)/(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)

We'll evaluate the limit:

lim (x^4 - x^3 + x^2 - x + 1)/(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) = [(-1)^4-(-1)^3+ (-1)^2 +1  + 1]/(1+1+1+1+1+1+1)

lim (x^4 - x^3 + x^2 - x + 1)/(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) = (1+1+1+1+1)/(1+1+1+1+1+1+1)

lim (x^4 - x^3 + x^2 - x + 1)/(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1) = 5/7

The value of the limit of the function, if x approaches to -1, is: lim (x^5+1)/(x^7+1) = 5/7.