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Q. `lim_(x->o)` ```(sin 3x + 7x) / (4x + sin2x)` = ?
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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`
Posted by llltkl on June 26, 2013 at 4:54 PM (Answer #1)
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