Which if the following statements are wrong
$S - I : \underset{x \to 0}{L t} \frac{e^{\frac{1}{x}} - e^{\frac{- 1}{x}}}{e^{\frac{1}{x}} + e^{\frac{- 1}{x}}} does not exist (\frac{e^{\frac{1}{x}} - e^{\frac{- 1}{x}}}{e^{\frac{1}{x}} + e^{\frac{- 1}{x}}})$
$S - II : If a \in R and f (x) = \{\begin{array}{cl} x^{a} (\frac{e^{\frac{1}{x}} - e^{\frac{- 1}{x}}}{e^{\frac{1}{x}} + e^{\frac{- 1}{x}}}) & , x \neq 0 \\ 0 & , x = 0 \end{array}$ is continuous $And differentiable \forall x \in R then a > 1$

Moderate

Solution

$\underset{x \to 0 +}{L t} \frac{1 - e^{\frac{- 2}{x}}}{1 + e^{\frac{- 2}{x}}} = \frac{1 - 0}{1 + 0} = 1, \underset{x \to 0 -}{L t} \frac{e^{\frac{- 2}{x}} - 1}{e^{\frac{- 2}{x}} + 1} = \frac{0 - 1}{0 + 1} = - 1$
$∴ S - I does not exists$
$\begin{array}{l} f (x) is a continuous at x = 0 ∴ f (0) = \lim_{_{x \to 0}} f (x) = \lim_{_{x \to 0}} x^{a} (\frac{e^{\frac{1}{x}} - e^{\frac{- 1}{x}}}{e^{\frac{1}{x}} + e^{\frac{- 1}{x}}}) \Rightarrow a > 0 \\ At x = 0 f (x) is differentiable \Rightarrow \lim_{x \to 0} \frac{f (x) - f (0)}{x - 0} exists \end{array}$
$\Rightarrow L_{x \to 0} x^{a - 1} (\frac{e^{\frac{1}{x}} - e^{\frac{- 1}{x}}}{e^{\frac{1}{x}} + e^{\frac{- 1}{x}}}) exists \begin{array}{l} \Rightarrow a - 1 > 0 \\ \Rightarrow a > 1 \end{array}$

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