Let f(x)=g(x)e1/x−e−1/xe1/x+e−1/x, where g is a continuous function then limx→0 f(x) exist if

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Let $f (x) = g (x) \frac{e^{1 / x} - e^{- 1 / x}}{e^{1 / x} + e^{- 1 / x}}$ , where g is a continuous function then $lim_{x \to 0} f (x)$ exist if

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$g (x) = x^{2} + 4$

$g (x) = xh (x), h (x)$ is polynomial

$g (x) = x + 2$

g (x) is a constant function

answer is C.

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Detailed Solution

$lim_{x \to 0^{+}} \frac{e^{1 / x} - e^{- 1 / x}}{e^{1 / x} + e^{- 1 / x}} = lim_{x \to 0^{+}} \frac{1 - e^{- 2 / x}}{1 + e^{- 2 / x}} = 1 lim_{x \to 0^{-}} \frac{e^{1 / x} - e^{- 1 / x}}{e^{1 / x} + e^{- 1 / x}} = lim_{x \to 0^{-}} \frac{e^{2 / x} - 1}{e^{2 / x} + 1} = - 1$
Hence $lim_{x \to 0} f (x) exists if g (x) = xh (x)$
Where h(x) is a continuous function. If $g (x) = a (a \neq 0)$ then
$lim_{x \to 0^{+}} f (x) = a, lim_{x \to 0^{-}} f (x) = - a$ . Thus $lim_{x \to 0} f (x)$ does not exist if $f (x) = x + 2$ or $x^{2} + 4$ or is a constant function.