# Q. `lim_(x->o)` ```(sin 3x + 7x) / (4x + sin2x)` = ?

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`lim_(x->0)(sin3x+7x)/(4x+sin2x)`

=`lim_(x->0)((sin3x)/x+7)/(4+(sin2x)/x)`

= `lim_(x->0)(3*(sin3x)/(3x)+7)/(4+2*(sin2x)/(2x))`

When `xrarr0` , both `3xrarr0` , and `2xrarr0 `

Therefore, the given limit becomes

= (3*1+7)/(4+2*1) (from limit rules, we know`lim_(u->0)(sinu)/u=1`

= 10/6

=5/3