hindi me anuvad kijie
translate into hindi
te yatha samayam vidhyalayam gachantu
please solve this question
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Answer:
ADWa
Explanation:
DAWadaDwadaddwadsdwaadasdwasdadasdwasdwadsdwas![\sqrt{x} \geq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \sqrt[n]{x} \sqrt[n]{x} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \alpha \pi \pi \pi](https://tex.z-dn.net/?f=%5Csqrt%7Bx%7D+%5Cgeq++%5Clim_%7Bn+%5Cto+%5Cinfty%7D+a_n+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%26amp%3B3%5C%5C4%26amp%3B5%26amp%3B6%5C%5C7%26amp%3B8%26amp%3B9%5Cend%7Barray%7D%5Cright%5D+%5Csqrt%5Bn%5D%7Bx%7D+%5Csqrt%5Bn%5D%7Bx%7D+%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26amp%3B2%26amp%3B3%5C%5C4%26amp%3B5%26amp%3B6%5C%5C7%26amp%3B8%26amp%3B9%5Cend%7Barray%7D%5Cright%5D+%5Calpha+%5Cpi+%5Cpi+%5Cpi "\sqrt{x} \geq \lim_{n \to \infty} a_n \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \sqrt[n]{x} \sqrt[n]{x} \left[\begin{array}{ccc}1&2&3\\4&5&6\\7&8&9\end{array}\right] \alpha \pi \pi \pi")
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