Mayli, keling, \(L\)-ga ega bo'lish uchun ildiz testidan boshlaylik.

\[L = \mathop {\lim }\limits_{n \to \infty } {\left| {\frac{{{n^{2n + 1}}}}{{{4^{3n}}}}{{\left( {2x + 17} \right)}^n}} \right|^{\frac{1}{n}}} = \mathop {\lim }\limits_{n \to \infty } \left| {\frac{{{n^{2 + \frac{1}{n}}}}}{{{4^3}}}\left( {2x + 17} \right)} \right| = \left| {2x + 17} \right|\mathop {\lim }\limits_{n \to \infty } \frac{{{n^{2 + \frac{1}{n}}}}}{{{4^3}}}\]

OK, biz shuni ko'ramizki, bu holda \(L\) \(x \ne - \frac{{17}}{2}\) bilan cheksiz bo'ladi va shuning uchun seriya \(x \ne - \frac{{17}}{2}\) uchun ajralib chiqadi. Biz shuningdek, quvvat seriyalari \(x = -\frac{{17}}{2}\) uchun birlashishini bilamiz (bu seriya uchun \(a\) qiymati!).

2-qadam

Shuning uchun biz yaqinlashuv oralig'i \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{17}}{2}}}\) va yaqinlashish radiusi \(\require{bbox} \bbox[2pt,border:1px solid black]{{R = 0}}\) ekanligini bilamiz.