1.

`(lim)_(x->0)(s i n a x)/(sinb x)a ,b ,!=0`

Answer» Here, we will use the property,
`lim_(x->0)sinx/x = 1`
`lim_(x->0) (sinax)/(sinbx) = lim_(x->0)((axsin ax)/(ax))/((bxsinbx)/(bx))`
`= lim_(x->0)(ax(sin ax)/(ax))/(bx(sin bx)/(bx))`
`= lim_(x->0) a/b[((sin ax)/(ax))/((sin bx)/(bx))]`
`= a/b[(lim_(x->0)((sin ax)/(ax)))/(lim_(x->0)((sin bx)/(bx)))]`
Now, let `ax = y and bx = z`
Then, our expression becomes,
`= a/b[(lim_(y->0)((sin y)/(y)))/(lim_(z->0)((sin z)/(z)))]`
`=a/b[1/1] = a/b`


Discussion

No Comment Found

Related InterviewSolutions