All limits below assume x-->1 (I am not writing it out to save some time).
By the sum and difference property of limits:
1. limx→a[f(x) + g(x)] = limx→a f(x) + limx→a g(x) ;
(the limit of a sum is the sum of the limits).
2. limx→a[f(x) − g(x)] = limx→a f(x) − limx→a g(x) ;
(the limit of a difference is the difference of the limits
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Lim [h(x) + j(x)] = 2, and Lim [h(x) - j(x)] = 1
<==>
lim h(x) + lim j(x) = 2
and
lim h(x) - lim j(x) = 1
Doing some simple algebra we get, lim h(x) = 2 - lim j(x) from the first limit expression. Substitute this into the second limit expression:
2 - lim j(x) - lim j(x) = 1 => 2- 2lim j(x) = 1 => -2 lim j(x) = -1 => lim j(x) = 1/2
Substitute back into our first expression to get:
lim h(x) = 3/2
Finally by the product property of limits
imx→a[f(x)g(x)] = limx→a f(x) · limx→a g(x);
(The limit of a product is the product of the limits)
we should be able to solve the limit.