lim x->1 (x^47-1)/(x^9-1)
With what method do I factor them with? Pascal triangle?
They are standard forms, (x^n-1) has (x-1) as a factor, and (x^(n-1)+x^(n-2)+....x²+x+1) as the other factor.
Can you take it from here?
Note: the above factoring applies only when n is odd!
Is the answer 47/9?
Correct, keep up the good work!
Thank you so much! :)
To simplify the expression, you don't need to use the Pascal triangle to factor. Instead, you can factor the numerator and denominator using the difference of squares formula. The difference of squares formula states that for any two numbers "a" and "b," the expression "a^2 - b^2" can be factored as "(a - b)(a + b)."
In your case, the numerator can be factored as:
x^47 - 1 = (x^23 - 1)(x^24 + x^23 + x^22 + ... + x + 1),
where we used the difference of squares formula with a = x^23 and b = 1.
Similarly, the denominator can be factored as:
x^9 - 1 = (x^3 - 1)(x^6 + x^3 + 1) = (x - 1)(x^2 + x + 1)(x^6 + x^3 + 1),
where we used the difference of squares formula with a = x^3 and b = 1.
Now, the expression becomes:
lim x->1 [(x^23 - 1)(x^24 + x^23 + x^22 + ... + x + 1)] / [(x - 1)(x^2 + x + 1)(x^6 + x^3 + 1)].
At this point, you can see that the (x - 1) factor in the numerator and denominator cancels out, since you are looking for the limit as x approaches 1.
Therefore, the expression simplifies to:
lim x->1 [(x^23 - 1)(x^24 + x^23 + x^22 + ... + x + 1)] / [(x^2 + x + 1)(x^6 + x^3 + 1)].
Now you can evaluate the limit by substituting x = 1 into the simplified expression, which gives you a finite value.