Thread: limit help

if lim_x->a[f(x) + g(x)] = 2, and lim_x->a[f(x) - g(x)] = 1, find lim_x->a[f(x)g(x)]


please explain how the answer is 3/4!!!!

$\displaystyle \lim_{x \to a} \left[f(x)+g(x)]^2 = 2^2$

$\displaystyle \lim_{x \to a} \left[f(x)^2 + 2f(x)g(x)+g(x)^2] = 4$

$\displaystyle \lim_{x \to a} \left[f(x)-g(x)]^2 = 1^2$

$\displaystyle \lim_{x \to a} \left[f(x)^2 - 2f(x)g(x)+g(x)^2] = 1$


$\displaystyle \lim_{x \to a} \left[f(x)^2 + 2f(x)g(x)+g(x)^2] -\lim_{x \to a} \left[f(x)^2 - 2f(x)g(x)+g(x)^2] = 4 - 1$

$\displaystyle \lim_{x \to a} \left[f(x)^2 + 2f(x)g(x)+g(x)^2] - \left[f(x)^2 - 2f(x)g(x)+g(x)^2] = 3$

$\displaystyle \lim_{x \to a} 4f(x)g(x) = 3$

$\displaystyle 4 \lim_{x \to a} f(x)g(x) = 3$

$\displaystyle \lim_{x \to a} f(x)g(x) = \frac{3}{4}$