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(A) We know that the fractional part function is defined as $\{y\} = y - [y]$.First,we evaluate the limit $\mathop {\lim }\limits_{x 	o {0^ + }} (1+x

Vedclass · Accepted Answer

$e^2 - 7$

Vedclass · Answer

(A) We know that the fractional part function is defined as $\{y\} = y - [y]$.First,we evaluate the limit $\mathop {\lim }\limits_{x 	o {0^ + }} (1+x)^{\frac{2}{x}}$.Using the standard limit formula $\mathop {\lim }\limits_{x 	o 0} (1+x)^{\frac{1}{x}} = e$,we have:$\mathop {\lim }\limits_{x 	o {0^ + }} (1+x)^{\frac{2}{x}} = \left( \mathop {\lim }\limits_{x 	o {0^ + }} (1+x)^{\frac{1}{x}} ight)^2 = e^2$.Since $e \approx 2.718$,$e^2 \approx 7.389$.As $x 	o 0^+$,$(1+x)^{\frac{2}{x}}$ approaches $e^2$ from the right side (since $(1+x)^{\frac{2}{x}}$ is an increasing function for small $x > 0$).Therefore,the greatest integer part $[(1+x)^{\frac{2}{x}}]$ will be $[e^2] = [7.389] = 7$.Thus,$\mathop {\lim }\limits_{x 	o {0^ + }} \left\{ (1+x)^{\frac{2}{x}} ight\} = \mathop {\lim }\limits_{x 	o {0^ + }} \left( (1+x)^{\frac{2}{x}} - [(1+x)^{\frac{2}{x}}] ight) = e^2 - 7$.

$\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {{{\left( {1 + x} \right)}^{\frac{2}{x}}}} \right\}$ is equal to (where $\{.\}$ denotes the fractional part of $x$)

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