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lim(xrarr0) (sinax)/(bx)...

`lim_(xrarr0) (sinax)/(bx)`

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Verified by Experts

`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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