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#