$\mathop {\lim }\limits_{n \to \infty } \left( {\frac{{{{\left( {n + 1} \right)}^{1/3}}}}{{{n^{4/3}}}} + \frac{{{{\left( {n + 2} \right)}^{1/3}}}}{{{n^{4/3}}}} + \dots + \frac{{{{\left( {2n} \right)}^{1/3}}}}{{{n^{4/3}}}}} \right)$ ની કિંમત શોધો.
- A$\frac{3}{4}(2^{4/3} - 1)$
- B$\frac{4}{3}(2^{3/4})$
- C$\frac{3}{4}(2^{4/3}) - \frac{4}{3}$
- D$\frac{4}{3}(2^{4/3})$
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$\mathop {\text{Lim}}\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^{4n} {\frac{{\sqrt n }}{{\sqrt r {{\left( {\,3\sqrt r + 4\sqrt n \,} \right)}^2}}}} $ ની કિંમત શોધો.
$a \in R, |a| > 1$ માટે,ધારો કે $\lim _{n \rightarrow \infty} \left( \frac{1+\sqrt[3]{2}+\ldots+\sqrt[3]{n}}{n^{7/3} \left( \frac{1}{(an+1)^2} + \frac{1}{(an+2)^2} + \ldots + \frac{1}{(an+n)^2} \right)} \right) = 54$. તો $a$ ની શક્ય કિંમત(ઓ) શોધો:
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$
$(1) 8$ $(2) -9$ $(3) -6$ $(4) 7$
$\lim _{n \rightarrow \infty} \frac{1}{n^3} \sum_{k=1}^n (k^2 x)$ ની કિંમત શોધો.
DifficultTS EAMCET 2004
View Solution$\lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n m}\right\}=$
$\int_{-1}^{2}(7 x-5) d x$ ની કિંમત સરવાળાના લક્ષ તરીકે મેળવો.
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