Functions Question 147
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{{e^{\alpha \ x}}-{e^{\beta \ x}}}{x}= $
[MP PET 1994; DCE 2005]
Options:
A) $ \alpha +\beta $
B) $ \frac{1}{\alpha }+\beta $
C) $ {{\alpha }^{2}}-{{\beta }^{2}} $
D) $ \alpha -\beta $
Show Answer
Answer:
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 . $
BETA
