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6 solutions
Very nice and simple solution!
Doesn't your solution presuppose x can't be 1?
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Yeh, but we are not evaluating at x=1 we are taking the limit, which means that x is "almost" 1.
Using L'Hôpital's Rule ,
x → 1 lim x 1 0 − 1 x 1 5 − 1 = x → 1 lim 1 0 x 9 1 5 x 1 4 = 1 0 1 5 = 1 . 5
Moderator note:
Simple standard approach.
One should remember to (demonstrate that they) check we have an indeterminate form before applying L'hopital.
You could always include a link for these rules in your solution for the viewers. Anyway, nice solution. Upvoted.
I will give a complicated solution! The numerator is the product of all cyclotomic polynomials which divide 15 and the denominator is all cyclotomic polynomias F d where d divides 10. So we get F 2 F 1 0 F 3 F 1 5 . But F p ( 1 ) = p if p is prime and 1 otherwise. So we get 2 3 .
I loved this solution! (+1)
Can you explain in full?
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When you have a non-determinated form like "0/0" you can derive numerator and denominator the limit will remain the same. This is De l'Hopital rule.
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So for this particular problem as x-->0, L'Hôpital's rule would not apply because there is no indeterminate form?
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@Scott Morris – That's correct. The limit as x --> 0 would evalute to -1/-1 or just 1. If you applied L'Hopitals rule your answer would be zero, which is incorrect.
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@Ahern Nelson – Thanks, Ahern Nelson
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@Scott Morris – Sorry I disagree when X->1 the limit goes to a 0/0 form than I can apply de l'Hopital
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@Angelo De Luca – Are we talking about the sa e exercise? No doubt for me
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@Angelo De Luca – The limit si for X->1 and the function is (x^15-1)/(x^10-1) than (1-1)/(1-1) = 0/0 indeterminate form therefore You can apply de l'Hopital
L'Hopital's rule will help in solving this question..
Substituting a = x 5 we have
x 1 0 − 1 x 1 5 − 1 = a 2 − 1 a 3 − 1 = ( a + 1 ) ( a − 1 ) ( a − 1 ) ( a 2 + a + 1 ) = a + 1 a 2 + a + 1
So
x → 1 lim x 1 0 − 1 x 1 5 − 1 = x → 1 lim x 5 + 1 x 1 0 + x 5 + 1 = 2 3