If the limit of f(x) as x tends to a = L and the limit of g(x) as x tends to a =M Proof that, lim[f(x)-g(x)] as x tends to a = L-M?

As it is written the statement is wrong.

1 Answer
Jan 24, 2018

We have to use the limit definition

Explanation:

Because of the limit definition we know:

lim_(x->c) h(x)=K iff given any epsilon > 0 there exists delta such that if 0 < |x-c| < delta => |h(x)-K| < epsilon

Now let's consider any epsilon > 0:

Since lim_(x->a) f(x)=L, given epsilon/2 > 0 there exists delta_1 such that if 0 < |x-a| < delta_1 => |f(x)-L| < epsilon/2

Since lim_(x->a) g(x)=M, given epsilon/2 > 0 there exists delta_2 such that if 0 < |x-a| < delta_2 => |g(x)-M| < epsilon/2

If we now take delta = min (delta_1, delta_2), we can say:

0<|x-a|< delta => 0<|x-a|< delta_1 and 0<|x-a|< delta_2

and therefore:

if 0<|x-a|< delta =>

|(f(x) -g(x))-(L-M)| < |(f(x) - L) - (g(x)-M)| <= |f(x) - L| +|g(x)-M| < epsilon/2 + epsilon/2=epsilon

That is, the lim_(x->a) [f(x) - g(x)] = L-M