Bài tập dãy số có giới hạn hữu hạn(1).

Question

Tính các giới hạn:
a)

$\mathop {\lim }\dfrac{\sqrt[3]{n^3+1}-3n}{\sqrt{n^2+n}+1}$ b)

$\mathop {\lim }\dfrac{2^n+3^n}{2^{n+1}+5.3^{n+1}}$

c)

$\mathop {\lim }\dfrac{2^n.3^{n+1}-5n}{6^n+5.3^n}$ d)

$\mathop {\lim }\dfrac{1+2+3+...+n}{2n^2-3}$

score 5 · Answer 1

a.

$\mathop {\lim }\dfrac{\sqrt[3]{n^3+1}-3n}{\sqrt{n^2+n}+1}=\lim\dfrac{\sqrt[3]{1+\dfrac{1}{n^3}}-3}{\sqrt{1+\dfrac{1}{n}}+\dfrac{1}{n}}=-2$
b.

$\mathop {\lim }\dfrac{2^n+3^n}{2^{n+1}+5.3^{n+1}}=\mathop {\lim}\dfrac{\left(\dfrac{2}{3}\right)^n+1}{2\left(\dfrac{2}{3}\right)^n+15}=\dfrac{1}{15}$
c.

$\mathop {\lim }\dfrac{2^n3^{n+1}-5n}{6^n+5.3^n}=\mathop {\lim }\dfrac{3-\dfrac{5n}{6^n}}{1+5\left(\dfrac{1}{2}\right)^n}=3$
d.

$\mathop {\lim }\dfrac{1+2+\ldots+n}{2n^2-3}=\mathop {\lim }\dfrac{n(n+1)}{4n^2-6}=\mathop {\lim }\dfrac{1+\dfrac{1}{n}}{4-\dfrac{6}{n^2}}=\dfrac{1}{4}$