Tìm các giới hạn b) lim (x đến 1) (x + x^2 + ... + x^n - n)/ (x - 1) c) lim (x đến 1) (x^100 - 2x + 1)/ (x^50 - 2x + 1) d) lim (x đến a) ( (x^n - a^n) - na^n-1 (x - a))/ (x - a)^2 Các bạn giúp mình bài 1.2 tính giới hạn ý b, c, d với

Đáp án:

$\begin{array}{l}
b)\mathop {\lim }\limits_{x \to 1} \dfrac{{x + {x^2} + ... + {x^n} - n}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{x - 1 + {x^2} - 1 + ... + {x^n} - 1}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right)\left( {1 + x + 1 + ... + {x^{n - 1}} + {x^{n - 2}} + ... + 1} \right)}}{{x - 1}}\\
 = \mathop {\lim }\limits_{x \to 1} {x^{n - 1}} + 2{x^{n - 2}} + ... + \left( {n - 1} \right)x + n\\
 = 1 + 2 + 3 + ... + n\\
 = \dfrac{{n\left( {n + 1} \right)}}{2}\\
c)\mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{100}} - 2x + 1}}{{{x^{50}} - 2x + 1}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {{x^{100}} - 1} \right) - 2\left( {x - 1} \right)}}{{\left( {{x^{50}} - 1} \right) - 2\left( {x - 1} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x - 1} \right).\left( {{x^{99}} + {x^{98}} + ... + x + 1 - 2} \right)}}{{\left( {x - 1} \right)\left( {{x^{49}} + {x^{48}} + ... + x + 1 - 2} \right)}}\\
 = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^{99}} + .. + x - 1}}{{{x^{49}} + ... + x - 1}}\\
 = \dfrac{{99 - 1}}{{49 - 1}} = \dfrac{{49}}{{24}}
\end{array}$

$\begin{array}{l}
d)\mathop {\lim }\limits_{x \to a} \dfrac{{\left( {{x^n} - {a^n}} \right) - n.{a^{n - 1}}\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to a} \dfrac{{\left( {x - a} \right)\left( {{x^{n - 1}} + {x^{n - 2}}.a + ... + {a^{n - 1}}} \right) - n.{a^{n - 1}}.\left( {x - a} \right)}}{{{{\left( {x - a} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to a} \dfrac{{\left( {x - a} \right)\left( {{x^{n - 1}} + {x^{n - 2}}.a + ... + {a^{n - 1}} - n.{a^{n - 1}}} \right)}}{{{{\left( {x - a} \right)}^2}}}\\
 = \mathop {\lim }\limits_{x \to a} \dfrac{{{x^{n - 1}} + {x^{n - 2}}.a + ... + x.{a^{n - 2}} - n.{a^{n - 1}}}}{{x - a}}\\
 = \mathop {\lim }\limits_{x \to a} \dfrac{{{x^{n - 1}} - {a^{n - 1}} + {x^{n - 2}}.a - {a^{n - 1}} + ... + x.{a^{n - 2}} - {a^{n - 1}}}}{{x - a}}\\
 = \mathop {\lim }\limits_{x \to a} \left[ \begin{array}{l}
\dfrac{{\left( {x - a} \right)\left( {{x^{n - 2}} + {x^{n - 3}}.a + ... + {a^{n - 2}}} \right)}}{{x - a}}\\
 + \dfrac{{\left( {x - a} \right)\left( {{x^{n - 3}}.a + {x^{n - 4}}.{a^2} + ... + {a^{n - 2}}} \right)}}{{x - a}}\\
 + ... + \dfrac{{{a^{n - 2}}\left( {x - a} \right)}}{{x - a}}
\end{array} \right]\\
 = \mathop {\lim }\limits_{x \to a} \left( \begin{array}{l}
{x^{n - 2}} + {x^{n - 3}}.a + ... + {a^{n - 2}}\\
 + {x^{n - 3}}.a + {x^{n - 4}}.{a^2} + ... + {a^{n - 2}}\\
 + ... + {a^{n - 2}}
\end{array} \right)\\
 = \left[ {\left( {n - 1} \right){a^{n - 2}} + \left( {n - 2} \right){a^{n - 2}} + ... + {a^{n - 2}}} \right]\\
 = {a^{n - 2}}.\dfrac{{\left( {n - 1} \right).n}}{2}\\
 = \dfrac{{{a^{n - 2}}.\left( {{n^2} - n} \right)}}{2}
\end{array}$