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lim(xrarr0) (sinax)/(sinbx),a,bne0...

`lim_(xrarr0) (sinax)/(sinbx),a,bne0`

Text Solution

Verified by Experts

The correct Answer is:
a/b
`underset(xrarr0)"lim"(sinax)/(sinbx)(a,bne0) ((0)/(0))`
`=underset(xrarr0)"lim"(sinax)/(ax).(bx)/(sinbx).(a)/(b)`
`=1xx1xx(a)/(b)(because underset(0rarr0)"lim"(sintheta)/(theta)=1)`
`=(a)/(b)`
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