Question

Suppose that `lim_(xrarrc) f(x) = 0` and there exists a constant `K` such that `∣g(x)∣ ≤ K " for all " x nec` in some open interval containing c. Show that# lim_(x→c) (f(x)g(x)) = 0#?

Answer 1

If `K = 0` then `g(x) = 0` for all `x!=c` in some open interval containing `c`, and so `lim_(x->c)(f(x)g(x)) = lim_(x->c)0=0`. Then, for the remainder of the explanation, we will assume `K > 0`.

Using the `epsilon-delta` definition of limits, we can say that `lim_(x->c)(f(x)g(x))=0` if for every `epsilon > 0` there exists a `delta > 0` such that `0<|x-c| < delta` implies `|f(x)g(x)-0| < epsilon`.

To show that this is the case, first we let `epsilon > 0` be arbitrary.

Now, as `K/epsilon>0` and `lim_(x->c)f(x)=0`, we know by the above definition there exists some `delta'>0` such that `|g(x)| < K` for `x in (c-delta',c+delta') \\\ {c}` and if `0<|x-c| < delta'` then `|f(x) - 0| < epsilon/K`.

Let `delta = delta'`. Then, for `0 < |x - c| < delta`, noting that as `|x-c| > 0` we know that `x!=c`, we have

`|f(x)g(x)-0| = |f(x)g(x)|`

`=|g(x)| * |f(x)|`

`<= K|f(x)|" "` (as `x in (c-delta',c+delta')\\\c => K >= |g(x)|`)

`< K(epsilon/K)" "` (as `0 < |x-c| < delta = delta' => |f(x)| < epsilon/K`)

`=epsilon`

Thus, as we have demonstrated the existence of such a `delta` for an arbitrary `epsilon`, we may conclude that `lim_(x->c)(f(x)g(x)) = 0`.

Answer 2

See below for a proof using the squeeze theorem.

Explanation:

`lim_(xrarrc)f(x) = 0` implies `lim_(xrarrc)abs(f(x)) = 0`

`abs(g(x)) <= K` if and only if `-K <= g(x) <= K` for `x` sufficiently close to `c` except perhaps at `x=c`.

`-K <= g(x) <= K` and `abs(f(x)) >= 0` implies

`-abs(f(x)) K <= abs(f(x)) g(x) <= abs(f(x)) K`.

Now, since `lim_(xrarrc)(-abs(f(x)) K) = 0 = lim_(xrarrc)(abs(f(x)) K)`,

we can conclude that `lim_(xrarrc)(abs(f(x)) g(x)) = 0`.