Question

The value of $\underset{x\to 0}{lim}(1-3x{)}^{\frac{1}{5x}}$ is?

• A. $e^{\frac{-3}{5}}$
• B. $e^{\frac{5}{3}}$
• C. $e^{\frac{-5}{3}}$
• D. $e^{-1}$

Answer 1

Answer: (A)

$e^{\frac{-3}{5}}$

Clearly form of the limit is $1^{\infty }$.

$Now,\underset{x\to 0}{lim}f(x{)}^{g\left(x\right)}=e^{(\underset{x\to 0}{lim}\left(f\right(x)-1).g\left(x\right))}$

$\therefore \underset{x\to 0}{lim}(1-3x{)}^{\frac{1}{5x}}=e^{\underset{x\to 0}{lim}(1-3x-1)\frac{1}{5x}}=e^{-\frac{3}{5}}$