A) g(x) is any constant function
B) g(x)=x
C) \[g(x)={{x}^{2}}\]
D) g(x) = x h (x), where h(x) is a polynomial.
Correct Answer: A
Solution :
[a] \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{{{e}^{1/x}}-{{e}^{-1/x}}}{{{e}^{1/x}}+{{e}^{-1/x}}}=\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\frac{1-{{e}^{-2/x}}}{1+{{e}^{-2}}^{/x}}=1\] and \[\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{e}^{1/x}}-{{e}^{-1/x}}}{{{e}^{1/x}}+{{e}^{-1/x}}}=\underset{x\to {{0}^{-}}}{\mathop{\lim }}\,\frac{{{e}^{2/x}}-1}{{{e}^{2/x}}+1}=-1\] Hence \[\underset{x\to 0}{\mathop{\lim }}\,f(x)\] exists if \[\underset{x\to 0}{\mathop{\lim }}\,\,g(x)=0\] If \[g(x)=a\ne 0\] (constant) then \[\underset{x\to {{0}^{+}}}{\mathop{\lim }}\,\,\,f(x)=a\] and \[\underset{x-{{0}_{-}}}{\mathop{\lim }}\,\,\,f(x)=-\,a\] Thus \[\underset{x\to 0}{\mathop{\lim }}\,\,f(x)\] doesn?t exist in this case.
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