Just Apply the L-Hospital rule twice

  by applying it once, u get

 Lim_{x-> 0}( - (m+2) sin(m+2)x  + msinx)/ (- (m+4)sin(m+4)x + (m+2) sin(m+2)x )

again applying L-Hospital,

Lim_{x-> 0}(-(m+2)²cos(m+2)x + m²cosmx )/ ( -(m+4)²cos(m+4)x + (m+2)²cos(m+2)x)

Putting limits, we get

(m² - (m+2)²)/( (m+2)² - (m+4)² )

=>  (m+1)/ ( m+3)

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