Question

${lim}_{x\to 0}\frac{1}{10}\frac{1-e^{-j5x}}{1-e^{-jx}}$ =

• A. 0
• B. 1.1
• C. 0.5
• D. 1

Answer 1

The correct option is C 0.5

Using L' Hospital rule

${lim}_{x\to 0}\frac{1-e^{-j5x}}{10(1-e^{-jx})}={lim}_{x\to 0}\frac{1}{10}\times \frac{5j{e}^{-j5x}}{j{e}^{-jx}}=\frac{5}{10}=0.5$