Question

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.

Answer 1

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`