\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\alpha \ x}}-{{e}^{\beta \ x}}}{x}=\] [MP PET 1994; DCE 2005]

\[{{\alpha }^{2}}-{{\beta }^{2}}\]

\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\alpha \ x}}-{{e}^{\b

1.	\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\alpha \ x}}-{{e}^{\beta \ x}}}{x}=\] [MP PET 1994; DCE 2005]
A.	\[\alpha +\beta \]
B.	\[\frac{1}{\alpha }+\beta \]
C.	\[{{\alpha }^{2}}-{{\beta }^{2}}\]
D.	\[\alpha -\beta \]
Answer» E.

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\[\underset{x\to 0}{\mathop{\lim }}\,\frac{{{e}^{\alpha \ x}}-{{e}^{\beta \ x}}}{x}=\] [MP PET 1994; DCE 2005]

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