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`lim_(xrarr0) (sinax)/(bx)` |
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Answer» `underset(Xrarr0)"lim"(sin ax)/(bx) ((0)/(0))` `=underset(xrarr0)"lim"(sinax)/(ax).(a)/(b)` `=1xx(a)/(b)=(a)/(b)(because underset(0rarr0)"lim"(sin theta)/(theta)=1)` |
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