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Evaluate, `lim_(xto0) (sin2x+3x)/(2x+tan3x)`.

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Given, `=underset(xto0)"lim"(sin2x+3x)/(2x+tan3x)=underset(xto0)"lim"((sin2x+3x)/(2x).2x)/((2x+tan3x)/(3x).3x)`
`=underset(xto0)"lim"(((sin2x)/(2x)+(3x)/(2x))2x)/(((2x)/(3x)+(tan3x)/(3x))3x)=underset(xto0)"lim"((sin2x)/(2x))/(2/3+(tan3x)/(3x)).2/3`
`=2/3underset(xto0)"lim"((sin2x)/(2x)+3/2)/(2/3+underset(xto0)"lim"(tan3x)/(3x))`
`=2/3((1+3/2)/(2/3+1))` `[therefore underset(xto0)"lim"(sinx)/x=1 " and " underset(xto0)"lim"(tanx)/x=1]`
`=2/3 xx (5/2)/(5/3)=2/3 xx 5/2 xx 3/5=1`

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