Show by means of an example that lim x→a [f(x) + g(x)] may exist even though neither lim x→a f(x) nor lim x→a g(x) exists.

Question

Show by means of an example that lim x→a [f(x) + g(x)]

 may exist even though neither lim x→a f(x)

 nor lim x→a g(x)

 exists. 

please explain with detail so that I understand, thanks.

Answer 1

this is a simplistic answer

f(x)=(4+x-a)/(x-a)

g(x)= -4/(x-a)

see f(x)+g(x)=1

x approach a from plus side lim f(x) is infinity, lim g(x) is minus infinity, lim f(x)+g(x)=1

x approach a from negative side lim f(x) is minus infinity, lim g(x) is infinity, lim f(x)+g(x)=1

Answer 2

Hi!

First, I want to briefly discuss the existence of a limit. Since I don't see a plus or minus in x→a I will assume we are working with a double sided limit. These are essentially said to exist if the single sided limits (approaching x=a from above and below) exist and are equal to each other. That is, lim x→a f(x) exists if lim x→a^- f(x) = lim x→a⁺ f(x).

So, with that in mind we want some functions that have a discontinuity at a when alone but do not when added together. The easiest example I can think of right now would be a pair of step functions (I've always denoted them as H(x-a) for Heaviside), which are defined as being 0 at x≤a and 1 at x>a. Let's look at f(x)=H(x-a) and g(x)=1-H(x-a). f will make a step up from 0 to 1 at a and g will make a step down from 1 to 0 at a. So that means lim x→a^- f(x)=lim x→a⁺ g(x)=0 and lim x→a⁺ f(x)=lim x→a^- g(x)=1, and since these one-sided limits are not equal neither lim x→a f(x) or lim x→a g(x) exist.

But, we can add these two step functions together, and since limits are linear we add the one-sided limits and find that in fact the two-sided limit does exist since both one-sided limits are equal to 1 when the f and g limits are combined. Since for the Heaviside function in particular one piece is less than or equal to, the sum of f and g is continuous and it should be safe to think of this case as simply a limit of 1 which is always 1 independent of the x value. However, for a discontinuous function you technically still need to think of it as a sum of limits rather than a limit of sums.

I hope this helps!