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

##### 2 Answers

It is not continuous at

#### Explanation:

If

Since

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

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

This leads is to an immediate problem as

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

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

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

**Method 2**

We can also the squeeze (or sandwich) theorem. We know

# -1 le sintheta le 1 #

Let

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

Multiply both sides by

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