Evaluate mathop {lim }limits_{x to {1^ + }} f | vedclass

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(B) Let $t = \{x\}$. As $x 	o 1^+$,$t 	o 0^+$.The expression becomes $\mathop {\lim }\limits_{t 	o 0^+} \frac{{(1 + t)^{1/t} - e^{1 - t/2}}}{1 - \c

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$\frac{{2e}}{3}$

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(B) Let $t = \{x\}$. As $x 	o 1^+$,$t 	o 0^+$.The expression becomes $\mathop {\lim }\limits_{t 	o 0^+} \frac{{(1 + t)^{1/t} - e^{1 - t/2}}}{1 - \cos t}$.Using the expansion $(1 + t)^{1/t} = e^{\frac{1}{t} \ln(1+t)} = e^{\frac{1}{t} (t - t^2/2 + t^3/3 - \dots)} = e^{1 - t/2 + t^2/3 - \dots} = e^{1 - t/2} \cdot e^{t^2/3 - \dots}$.Substituting this into the limit:$\mathop {\lim }\limits_{t 	o 0^+} \frac{e^{1 - t/2} (e^{t^2/3} - 1)}{t^2/2} = \mathop {\lim }\limits_{t 	o 0^+} \frac{e^{1 - t/2} (1 + t^2/3 + \dots - 1)}{t^2/2}$.$= \mathop {\lim }\limits_{t 	o 0^+} \frac{e^{1 - t/2} \cdot t^2/3}{t^2/2} = \frac{e}{3} \cdot 2 = \frac{2e}{3}$.

Evaluate $\mathop {\lim }\limits_{x \to {1^ + }} \frac{{{{\left( {1 + \left\{ x \right\}} \right)}^{\frac{1}{{\left\{ x \right\}}}}} - \frac{e}{{\sqrt {{e^{\left\{ x \right\}}}} }}}}{{1 - \cos \left\{ x \right\}}}$ (where $\{.\}$ denotes the fractional part function).

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