could you find the limit of lim x to 0 of the function x/((6/(x+4))-(3/(x+2)))

To find the limit \[ \lim_{x \to 0} \frac{x}{\left( \frac{6}{x+4} \right) - \left( \frac{3}{x+2} \right)} \] we start by simplifying the denominator. The expression in the denominator is: \[ \frac{6}{x+4} - \frac{3}{x+2} \] To combine these fractions, we need a common denominator, which is \((x+4)(x+2)\): \[ \frac{6(x+2) - 3(x+4)}{(x+4)(x+2)} \] Now, simplifying the numerator: \[ 6(x+2) = 6x + 12 \] \[ 3(x+4) = 3x + 12 \] Thus, the numerator becomes: \[ 6x + 12 - (3x + 12) = 6x + 12 - 3x - 12 = 3x \] Now we have: \[ \frac{6}{x+4} - \frac{3}{x+2} = \frac{3x}{(x+4)(x+2)} \] Substituting this result back into our limit, we have: \[ \lim_{x \to 0} \frac{x}{\frac{3x}{(x+4)(x+2)}} \] This expression simplifies to: \[ \lim_{x \to 0} \frac{x (x+4)(x+2)}{3x} \] If \(x \neq 0\), we can cancel \(x\) in the numerator and the denominator: \[ \lim_{x \to 0} \frac{(x+4)(x+2)}{3} \] Now we can substitute \(x = 0\): \[ \frac{(0 + 4)(0 + 2)}{3} = \frac{4 \cdot 2}{3} = \frac{8}{3} \] Thus, the limit is: \[ \boxed{\frac{8}{3}} \]