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Originally Posted by luiza001 You can use l'hopital's rule twice or factor out x-1 from the the numerator and denominator.
Originally Posted by JaguarXJS You can use l'hopital's rule twice or factor out x-1 from the the numerator and denominator. Actually just apply L'Hopital once (or factor out the $(x-1)$ as suggested).
Originally Posted by JaguarXJS You can use l'hopital's rule twice or .... You cannot use it twice, because after the first use, the expression is no longer an indeterminate form. Originally Posted by JaguarXJS ... factor out x-1 from the numerator and denominator. Can you do the factorization without involving (read: writing out) dozens of terms in both the numerator and denominator?
Last edited by greg1313; Apr 4th 2018 at 07:32 AM.
Originally Posted by greg1313 You cannot use it twice, because after the first use, the expression is no longer an indeterminate form. That is the whole point of l'Hôpital's rule. $\displaystyle{{\lim _{x \to 1}}\dfrac{{{x^{100}} - 2x + 1}}{{{x^{50}} - 2x + 1}}\mathop = \limits^H {\lim _{x \to 1}}\dfrac{{100{x^{99}} - 2}}{{50{x^{49}} - 2}}} = ?$