By the definition of continuity, how do you show that #xsin(1/x)# is continuous at x=0?

2 Answers
May 16, 2017

Answer:

It is not continuous at #0#.

Explanation:

#f# is continuous at #a# if and only if #lim_(xrarra)f(x) = f(a)#

If #f(a)# does not exist, then #f# is not continuous at #a#.

Since #0sin(1/0)# does not exists, #xsin(1/x)# is not continuous at #0#.

May 16, 2017

Answer:

The function as defined is not, but a slight modification is if we define a modified function by:

# f(x) = { ( xsin(1/x), x != 0), (0,x=0):} #

Explanation:

We will need the definition of continuity which is that:

# f(x)# is continuous at #x=a iff lim_(x rarr a)f(x)=f(a) #

So, in order to prove that the function defined by:

# f(x) = xsin (1/x) #

Is continuous at #x=0# we must show that

# lim_(x rarr 0)xsin(1/x) = f(0) #

This leads is to an immediate problem as #f(0)# is clearly undefined.

This is, however, not the end of the problem. If we look at the graph of the function, it certainly looks as though the function #f(x)# is continuous:

graph{xsin(1/x) [-0.2034, 0.2068, -0.1015, 0.1036]}

Firstly, Let us try and establish if the above limit exists

We can very easily show the limit exists and find its value:

Method 1:

Let #z = 1/x# then as # x rarr 0 => z rarr oo#

So then, the limit can be written:

# lim_(x rarr 0)xsin(1/x) = lim_(z rarr oo) (1/z)sinz #
# " " = lim_(z rarr oo) (sinz)/z #
# " " = 0 #

As #|sin(z)| le 1# and #1/z rarr 0# as #z rarr oo#

Method 2
We can also the squeeze (or sandwich) theorem. We know #sinx# oscillates between #+-1#, that is:,

# -1 le sintheta le 1 #

Let #theta = 1/x# then

# -1 le sin(1/x) le 1 #

Multiply both sides by #x#, then depending upon the sign of #x# we have:

# x >0 => -x <= xsin(1/x) <= x#
# x<0 => -x >= xsin(1/x) >= x#

And since:

# lim_(x rarr 0^+) x = lim_(x rarr 0^+) -x = 0#
# lim_(x rarr 0^-) x = lim_(x rarr 0^-) -x = 0#

We can use the squeeze theorem to say

# lim_(x rarr 0) xsin(1/x) = 0#

So how does this help us? We have shown the limit exists, but we still have the issue with #f(0)# being undefined. We can easily deal with this by modifying the function to explicitly define #f(0)=0# as follows:

# f(x) = { ( xsin(1/x), x != 0), (0,x=0):} #