A) \[\alpha +\beta \]
B) \[\frac{1}{\alpha }+\beta \]
C) \[{{\alpha }^{2}}-{{\beta }^{2}}\]
D) \[\alpha -\beta \]
Correct Answer: D
Solution :
\[\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\alpha x}}-{{e}^{\beta x}}}{x}=\underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\alpha x}}-1-{{e}^{\beta x}}+1}{x}\] \[=\alpha \underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\alpha x}}-1}{\alpha x}-\beta \underset{x\to 0}{\mathop{\lim }}\,\,\frac{{{e}^{\beta x}}-1}{\beta x}\]\[=\alpha .1-\beta .1=\alpha -\beta .\]
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