The limit
lim as x approaches 1 of (x^12-1)/(x-1)
represents the derivative of some function f(x) at some number a. Find f and a.
derivative of top = 12 x^11
derivative of bottom = 1
so I get 12
If f = 12 x
f'(x) = 12
for any old x = a
To find the function f(x) and the number a, we need to use the concept of the derivative.
The limit you provided, lim as x approaches 1 of (x^12-1)/(x-1), can be simplified using algebraic manipulation.
Step 1: Simplify the expression by factoring the numerator as a difference of squares:
(x^12 - 1) = (x^6 + 1)(x^6 - 1)
(x^6 + 1) can be further factored as a sum of cubes using the formula (a^3 + b^3) = (a + b)(a^2 - ab + b^2):
(x^6 + 1) = (x^2 + 1)(x^4 - x^2 + 1)
Step 2: Simplify the expression by factoring the denominator using the difference of squares:
(x - 1) = (sqrt(x) - 1)(sqrt(x) + 1)
The simplified expression becomes:
lim as x approaches 1 of [(x^2 + 1)(x^4 - x^2 + 1)] / [(sqrt(x) - 1)(sqrt(x) + 1)]
Now, let's find the function f(x) and the number a.
The given limit can be rewritten using the concept of the derivative:
lim as x approaches a of (f(x) - f(a)) / (x - a), where f(x) is the function we want to find and a is the number we want to determine.
Comparing this form with the given limit, we can match it with:
lim as x approaches 1 of [(x^2 + 1)(x^4 - x^2 + 1)] / [(sqrt(x) - 1)(sqrt(x) + 1)] = (f(x) - f(1)) / (x - 1)
We can see that a = 1.
To find f(x), we need to simplify the numerator and denominator of the fraction obtained by comparing the limits.
Numerator:
(x^2 + 1)(x^4 - x^2 + 1) = x^6 + x^4 + x^2 + x^4 - x^2 + 1 - x^2 - 1 + 1
Simplifying further, the numerator becomes: x^6 + 2x^4
Now, we can rewrite the limit as:
lim as x approaches 1 of (x^6 + 2x^4) / [(sqrt(x) - 1)(sqrt(x) + 1)] = (f(x) - f(1)) / (x - 1)
Comparing the limits, we can see that f(x) = x^6 + 2x^4.
Therefore, the function f(x) is f(x) = x^6 + 2x^4 and the number a is a = 1.