Bài tập 25 trang 108 sgk Toán 11 tập 2 KNTT: Tính các giới hạn sau:
a) $\lim_{n\rightarrow + \infty} \frac{1+3+5+...+(2n-1)}{n^{2}+2n+3}$
b) $\lim_{n\rightarrow + \infty} (1+ \frac{2}{3}+ \frac{4}{9}+...+\frac{2^{n}}{3^{n}})$
c) $\lim_{x\rightarrow + \infty} \frac{2x^{2}+3x-2}{x^{2}-4 }$
d) $\lim_{x\rightarrow + \infty} ( \sqrt {4x^{2}+x+1}+2x)$
Bài Làm:
a) $\frac{1+3+5+...+(2n-1)}{n^{2}+2n+3} = \frac{n}{n^{2}+2n+3}$
Khi $n \to \infty$, ta có:
$\lim_{n\rightarrow + \infty} \frac{n}{n^{2}+2n+3} = \lim_{n\rightarrow + \infty} \frac{1}{n+2+\frac{3}{n}} = \frac{1}{\infty} = 0$
Vậy, $\lim_{n\rightarrow + \infty} \frac{1+3+5+...+(2n-1)}{n^{2}+2n+3} = 0$.
b) Ta có:
$1+ \frac{2}{3}+ \frac{4}{9}+...+\frac{2^{n}}{3^{n}} = \sum_{k=0}^{n} \frac{2^{k}}{3^{k}}$
Khi $n \to \infty$, ta có:
$\lim_{n\rightarrow + \infty} \sum_{k=0}^{n} \frac{2^{k}}{3^{k}} = \sum_{k=0}^{\infty} \frac{2^{k}}{3^{k}} = \sum_{k=0}^{\infty} \left(\frac{2}{3}\right)^{k} = \frac{1}{1 - \frac{2}{3}} = \frac{1}{\frac{1}{3}} = 3$
Vậy, $\lim_{n\rightarrow + \infty} (1+ \frac{2}{3}+ \frac{4}{9}+...+\frac{2^{n}}{3^{n}}) = 3$.
c) $\frac{2x^{2}+3x-2}{x^{2}-4} = \frac{x^{2}(2 + \frac{3}{x} - \frac{2}{x^{2}})}{x^{2}(1 - \frac{4}{x^{2}})}$
Khi $x \to \infty$, ta có:
$\lim_{x\rightarrow + \infty} \frac{2x^{2}+3x-2}{x^{2}-4} = \lim_{x\rightarrow + \infty} \frac{2 + \frac{3}{x} - \frac{2}{x^{2}}}{1 - \frac{4}{x^{2}}} = \frac{2 + 0 - 0}{1 - 0} = 2$
Vậy, $\lim_{x\rightarrow + \infty} \frac{2x^{2}+3x-2}{x^{2}-4} = 2$.
d) $\sqrt {4x^{2}+x+1}+2x = \sqrt {x^{2}(4 + \frac{1}{x} + \frac{1}{x^{2}})}+2x $
$= \sqrt {x^{2}\left(4 + \frac{1}{x} + \frac{1}{x^{2}}\right)}+2x$
Khi $x \to \infty$, ta có:
$\lim_{x\rightarrow + \infty} \sqrt {x^{2}\left(4 + \frac{1}{x} + \frac{1}{x^{2}}\right)}+2x$
$= \lim_{x\rightarrow + \infty} x \sqrt {4 + \frac{1}{x} + \frac{1}{x^{2}}}+2x$
$= \infty \times (4 + 0 + 0) + \infty = \infty$
Vậy, $\lim_{x\rightarrow + \infty} ( \sqrt {4x^{2}+x+1}+2x) = \infty$.
