# Calculate limit with special limits limit (sinx+5sin3x)/(sin2x+2sin4x) x-->0

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Since the equation of function whose limit you need to evaluate contains the function `sin (f(x))` , yields that you may use the following special limit, such that:

`lim_(f(x)->0) (sinf(x))/(f(x)) = 1`

Reasoning by analogy, you may create the special limit to each member of equation, such that:

`lim_(x) ((sin x)/x*x+5(sin3x)/(3x)*3x)/(2x*(sin2x)/(2x)+2*4x*(sin4x)/(4x))`

Using the distributive property of limits, yields:

`(lim_(x->0)(sin x)/x*x+5lim_(x->0)(sin3x)/(3x)*3x)/(2x*lim_(x->0)(sin2x)/(2x)+2*4x*lim_(x->0)(sin4x)/(4x)) = lim_(x->0)(x + 15x)/(4x + 8x)`

`lim_(x->0)(x + 15x)/(x + 8x) = lim_(x->0)(16x)/(10x)`

Reducing duplicate factors, yields:

`lim_(x->0)(x + 15x)/(2x + 8x) = 8/5`

**Hence, evaluating the given limit, using the special limit `lim_(f(x)->0) (sinf(x))/(f(x)) = 1` , yields**` lim_(x) ((sinx)+5(sin3x))/((sin2x)+2(sin4x)) = 8/5.`