Cho \[{u_n} = \frac{{1 - 4n}}{{5n}}\]. Khi đó \[lim\,{u_n}\]bằng?

A.\[\lim {q^n} = 0\]

B. \[\lim q = 0\]

C. \[\lim \left( {n.q} \right) = 0\]

D. \[\lim \frac{n}{q} = 0\]

A.\[\lim \left( {{u_n} + {v_n}} \right) = L + M\]

B. \[\lim \left( {{u_n} + {v_n}} \right) = L - M\]

C. \[\lim \left( {{u_n} - {v_n}} \right) = L + M\]

D. \[\lim \left( {{u_n} - {v_n}} \right) = L.M\]

A.\[\frac{{{a^2}\left( {{2^{100}} - 1} \right)}}{{{2^{100}}}}\]

B. \[2{a^2}\]

C. \[\frac{{{a^2}}}{{{2^{100}}}}\]

D. \[\frac{{{a^2}\left( {{2^{99}} - 1} \right)}}{{{2^{98}}}}\]

A.\[\lim \sqrt[3]{{{u_n}}} = L\]

B. \[\lim \sqrt {{u_n}} = L\]

C. \[\lim \sqrt {{u_n}} = \sqrt L \]

D. \[\lim \sqrt[3]{{{u_n}}} = \sqrt[3]{L}\]

A.\[\lim \frac{{3{u_n} - 1}}{{{u_n} + 1}} = 3\]

B. \[\lim \frac{{3{u_n} - 1}}{{{u_n} + 1}} = - 1\]

C. \[\lim \frac{{3{u_n} - 1}}{{{u_n} + 1}} = 2\]

D. \[\lim \frac{{3{u_n} - 1}}{{{u_n} + 1}} = 1\]

A.\[\frac{1}{5}\]

B. \(\frac{1}{2}\)

C. 0

D. \[\frac{{ - 3}}{2}\]