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  • Stuck here, help me understand: - Limit , continuity and differentiability - JEE Main-9

\lim_{n\rightarrow \infty }(1/n^{3}+2/n^{3}+3/n^{3}+....+ n/n^{3})  equals 

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As we have learned

The SANDWICH THEOREM -

If \:\:f(x)\leq g(x)\leq h(x)\:\:for\:every\:(x)\:in\:the\:deleted\:neighbourhood\:of\:(a).


and\:\:\:\lim_{x\rightarrow a}\:\:f(x)=\lim_{x\rightarrow a}\:\:h(x)=l


Then\;\:\lim_{x\rightarrow a}\:g(x)\:\:is\:also\:equal\:to\:l

- wherein

 

Where\:\:\:x\epsilon \:(a-\delta ,\:a+\delta )\:\:and\:\:\delta\: \:is\:very\:small.

 

 once replacing all terms by 1/n^{3} then sum becomes n/n^{3}=\frac{1}{n^{2}}

once replacing all terms by \frac{n}{n^{3}}  then the sum becomes \frac{n^{2}}{n^{3}}=1/n  , we can write that 

1/n^{3}+1/n^{3}+1/n^{3}+....+1/n^{3}< 1/n^{3}+2/n^{3}+....+n/n^{3}< n/n^{3}+n/n^{3}+...+n/n^{3}

\Rightarrow 1/n^{2}< 1/n^{3}+2/n^{3}+...+n/n^{3}< 1/n

taking \lim_{n\rightarrow \infty } throughtout ,  \lim_{n\rightarrow \infty }1/n^{2}=\lim_{n\rightarrow \infty } 1/n=0

\therefore \lim_{n\rightarrow \infty } \left ( 1/n^{3}+2/n^{3}+....+n/n^{3} \right )=0(USING SANDWICH THEOREM ) 

 

 

 

 

 


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