Question
Let $f(x) = \lim_{\theta \rightarrow 0}\mspace{2mu}\left( \frac{cos\pi x - x^{\left( \frac{2}{\theta} \right)}sin(x - 1)}{1 + x^{\left( \frac{2}{\theta} \right)}(x - 1)} \right),x \in R$.
Consider the following two statements :
(I) $f(x)$ is discontinous at $x = 1$.
(II) $f(x)$ is continous at $x = - 1$.
Then,
Options
Solution
$f(x) = \left\{ \begin{matrix} cos\pi x & x \rightarrow 1^{-} \\ \frac{- sin(x - 1)}{(x - 1)} & x \rightarrow 1^{+} \end{matrix} \right.\ $
$${RHL = \lim_{x \rightarrow 1}\mspace{2mu}\frac{- sin(x - 1)}{(x - 1)} = - 1 }{LHL = \lim_{x \rightarrow 1}\mspace{2mu} cos\pi x = - 1,f(1) = - 1 }$$$f(x)$ is continuous at $x = 1$
$${f(x) = \left\{ \begin{matrix} \frac{- sin(x - 1)}{- (x - 1)} & x \rightarrow - 1^{-} \\ cos\pi x & x \rightarrow - 1^{+} \end{matrix} \right.\ }{RHL = \lim_{x \rightarrow - 1}\mspace{2mu} cos\pi x = - 1 }{LHL = \lim_{x \rightarrow - 1}\mspace{2mu}\frac{- sin(x - 1)}{(x - 1)} = \frac{sin2}{- 2} }$$$f(x)$ is discontinuous at $x = - 1$
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