MCQ
$\mathop {\lim }\limits_{x \to {0^ + }} \left\{ {{{\left( {1 + x} \right)}^{\frac{2}{x}}}} \right\}$ is equal to (where $\{.\}$ denotes the fractional part of $x$)
- ✓$e^2 -7$
- B$e^2 -8$
- C$e^2 -6$
- D$7$
And $\lim _{x \rightarrow 0}(1+x)^{\frac{2}{x}}=\exp \lim _{x \rightarrow 0}\left(\frac{2}{x} \log (1+x)\right)=\exp \lim _{x \rightarrow 0}\left(\frac{2 \frac{1}{1+x}}{1}\right)=e^{2}$
since$\left[\lim _{x \rightarrow \infty} f(x)^{g(x)}=e^{\left(\lim _{x \rightarrow \infty} g(x) \cdot \ln f(x)\right)}\right]$
$\therefore \lim _{x \rightarrow 0}\left\{(1+x)^{\frac{2}{x}}\right\}=e^{2}-\left[e^{2}\right]=e^{2}-7$
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