How do you prove that the function x sin ( 1 x ) is continuous at x=0?

Question

How do you prove that the function       x          sin              (                  1          x                )             is continuous at x=0?

HadoHaurrysap3w · Accepted Answer

The function, as given, is not continuous at 0 as

0 \sin (\frac{1}{0})

is not defined. However, we may make a slight modification to make the function continuous, defining f(x) as

f (x) = {\begin{cases} x \sin (\frac{1}{x}) if x \neq 0 \\ 0 if x = 0 \end{cases}

We will proceed using this modified function.
Using the

ε - δ

definition of a limit, we must show that for any

ε > 0

there exists a

δ > 0

such that if

0 < | x - 0 | < δ

then

| f (x) - f (0) | < ε

To do so, we first let

ε > 0

be arbitrary. Next, let

δ = \frac{ε}{2}

. Now, suppose

0 < | x - 0 | = | x | < δ

. Note that as

| x | > 0

we have

f (x) = x \sin (\frac{1}{x})

. Proceeding,

| f (x) - f (0) | = | x \sin (\frac{1}{x}) - 0 |

= | x \sin (\frac{1}{x}) |

= | x | \cdot | \sin (\frac{1}{x}) |

\leq | x |

(as

| \sin (\frac{1}{x}) | \leq 1

for all

x \neq 0

)

< δ

= \frac{ε}{2}

< ε

So, as for an arbitrary

ε > 0

there exists a

δ > 0

such that if

0 < | x - 0 | < δ

then

| f (x) - f (0) | < ε

, we have shown that f(x) is continuous at x=0.

How do you prove that the function xsin(1/x) is continuous at x=0?

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