Q. 19 - Calculus

Question

$0^0$ that approaches (a) $1,$(b) $0,$ (c) $\infty .$

Step-by-Step Solution

Answer

$0^0$ that approaches to (a) $1.$

Step 1. Given information

The given expression is $0^0,$ we need to write $0^0$ approaches to (a) $1,$ (b) $0,$ (c) $\infty .$

Step 2. The explanation for the correct option

For example, $\underset{x\to 0^{+}}{\lim }{x}^x$

                  $=\underset{x\to 0}{\lim }{e}^{\ln z^x}$

                   $=\underset{x\to 0}{\lim }{e}^{x\ln x}$

                  $=\underset{x\to 0}{\lim }{e}^{\frac{\ln x}{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}}$

                  $=\underset{x\to 0}{\lim }\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.}}{-{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x^2$}\right.}}$

                  $=\underset{x\to 0}{\lim }{e}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$x$}\right.·-x^2}$

                  $=\underset{x\to 0}{\lim }{e}^{-x}$

                  $=e^0$

                   $=1$