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#?

Question

William Abdalla · Accepted Answer

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#.

Charles Anctil · Answer

See below for a proof using the squeeze theorem.

#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#.

Emilia Aguilar · Answer

undefined To show that lim_(x→c) (f(x)g(x)) = 0, we can use the limit properties and the given information.

Since lim_(x→c) f(x) = 0, we know that as x approaches c, the function f(x) approaches 0.

We are also given that there exists a constant K such that ∣g(x)∣ ≤ K for all x nec in some open interval containing c. This means that the absolute value of g(x) is always less than or equal to K in that interval.

Now, let's consider the product f(x)g(x). As x approaches c, both f(x) and g(x) approach 0 (from the given information).

Since the absolute value of g(x) is always less than or equal to K, we can say that ∣f(x)g(x)∣ ≤ K * ∣f(x)∣ for all x nec in the open interval containing c.

Since f(x) approaches 0 as x approaches c, we can say that ∣f(x)∣ approaches 0 as well.

Therefore, as x approaches c, ∣f(x)g(x)∣ approaches 0 * K = 0.

Hence, we have shown that lim_(x→c) (f(x)g(x)) = 0.

Axel Alvarez · Answer

undefined Given that $ \lim_{x \to c} f(x) = 0 $ and $ |g(x)| \leq K $ for all $ x $ near $ c $, we need to show that $ \lim_{x \to c} (f(x)g(x)) = 0 $. Since $ \lim_{x \to c} f(x) = 0 $, for any $ \varepsilon > 0 $, there exists a $ \delta_1 > 0 $ such that $ |f(x)| < \varepsilon $ whenever $ 0 < |x - c| < \delta_1 $. Similarly, since $ |g(x)| \leq K $ for all $ x $ near $ c $, for the same $ \varepsilon > 0 $, there exists a $ \delta_2 > 0 $ such that $ |g(x)| \leq K $ whenever $ 0 < |x - c| < \delta_2 $. Now, consider $ |f(x)g(x)| $. We know that $ |f(x)| < \varepsilon $ and $ |g(x)| \leq K $ when $ 0 < |x - c| < \min(\delta_1, \delta_2) $. Therefore, $ |f(x)g(x)| < \varepsilon K $ for $ 0 < |x - c| < \min(\delta_1, \delta_2) $. Since $ \varepsilon $ can be made arbitrarily small, and $ K $ is a constant, $ \varepsilon K $ approaches 0 as $ \varepsilon $ approaches 0. Hence, $ \lim_{x \to c} (f(x)g(x)) = 0 $.

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#?