A) 1
B) \[{{e}^{b-a}}\]
C) \[{{e}^{a-b}}\]
D) \[{{e}^{b}}\]
Correct Answer: C
Solution :
\[\underset{x\to \infty }{\mathop{\text{lim}}}\,\,{{\left( \frac{x+a}{x+b} \right)}^{x+b}}=\underset{x\to \infty }{\mathop{\text{lim}}}\,\,{{\left( 1+\frac{a-b}{x+b} \right)}^{x+b}}\] \[=\underset{x\to \infty }{\mathop{\text{lim}}}\,\,{{\left\{ {{\left( 1+\frac{a-b}{x+b} \right)}^{\frac{x+b}{a-b}}} \right\}}^{a-b}}={{e}^{a-b}}\].
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