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`lim_(h rarr 0) [(1)/(h.root(3)(8+h))-(1)/(2h)]` का मान क्या होगा ?A. `(1)/(48)`B. `-(1)/(48)`C. `(1)/(18)`D. `(1)/(38)` |
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Answer» Correct Answer - B `underset(hrarr0)lim[(1)/(hroot(3)(8+h))-(1)/(2h)]=underset(hrarr0)lim[(2-root(3)(8+h))/(2hroot(3)(8+h))]` `=underset(hrarr0)lim[(2{1-(1+h//8)^(1//3)})/(2hroot(3)(8+h))]=underset(hrarr0)lim(2{1-1-(1)/(3).(h/8)+"..."})/(2hroot(3)(8+h))` `=underset(hrarr0)lim (-1//24+h"की उच्च घातों को छोड़ने पर ")/(root(3)(8+h))=-(1//24)/(2)=-(1)/(48)` |
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