Question

: Discuss the continuity of the function f(x)=xtanh(1/x) at x = 0 given that x≠0 and f(0)=0.

Answer 1

Explanation:

Given that:

`f(x) = x tan ( (1)/(x))`

For continuity;
`\lim \limits_(x \to 0^-) f(x) = \lim \limits_(x \to 0^+) f(x) = f(0)`

Now;

`\lim \limits_(x \to 0^-) f(x) = \lim \limits_(x \to 0^-) xtan ((1)/(x))`

`= \lim \limits_(x \to 0^-) (tan ((1)/(x)))/((1)/(x))`

`= \lim \limits_(x \to 0^-) (tan ((1)/(0)))/((1)/(0))`

`= (\infty)/(\infty) \to intermediate \ form`

Using L'Hospital Rule

`\lim \limits _(x \to 0^-) [ x \ tan ((1)/(x))] = \lim \limits _(x \to 0^-) ((d)/(dx) \ tan (1)/(x) )/((d)/(dx) (1)/(x) )`

`= \lim \limits _(x \to 0^-) (sec^(1) ((1)/(x)) * (-1)/(x^2) )/((-1)/(x^2) )`

`= \lim \limits _(x \to 0^-) sec^(1) ((1)/(x))`

`= sec^1((1)/(0))`

`=sec^1 (\infty) \to` not defined

∴

`\lim \limits _(x \to 0^- )[x \ tan ((1)/(x))] \\e 0 = f(0)`

Hence, f(x) is not continuous.