Answer:
Step-by-step explanation:
Given that:
[tex]f(x) = x tan ( \dfrac{1}{x})[/tex]
For continuity; [tex]\lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^+} f(x) = f(0)[/tex]
Now;
[tex]\lim \limits_{x \to 0^-} f(x) = \lim \limits_{x \to 0^-} xtan (\dfrac{1}{x})[/tex]
[tex]= \lim \limits_{x \to 0^-} \dfrac{tan (\dfrac{1}{x})}{\dfrac{1}{x}}[/tex]
[tex]= \lim \limits_{x \to 0^-} \dfrac{tan (\dfrac{1}{0})}{\dfrac{1}{0}}[/tex]
[tex]= \dfrac{\infty}{\infty} \to intermediate \ form[/tex]
Using L'Hospital Rule
[tex]\lim \limits _{x \to 0^-} [ x \ tan (\dfrac{1}{x})] = \lim \limits _{x \to 0^-} \dfrac{\dfrac{d}{dx} \ tan \dfrac{1}{x} }{\dfrac{d}{dx} \dfrac{1}{x} }[/tex]
[tex]= \lim \limits _{x \to 0^-} \dfrac{sec^{1} (\dfrac{1}{x}) * \dfrac{-1}{x^2} }{\dfrac{-1}{x^2} }[/tex]
[tex]= \lim \limits _{x \to 0^-} sec^{1} (\dfrac{1}{x})[/tex]
[tex]= sec^1(\dfrac{1}{0})[/tex]
[tex]=sec^1 (\infty) \to[/tex] not defined
∴
[tex]\lim \limits _{x \to 0^- }[x \ tan (\dfrac{1}{x})] \ne 0 = f(0)[/tex]
Hence, f(x) is not continuous.
