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$\lim\limits_{x\to 0}\large\frac{\sin x}{x}$


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$\lim\limits_{x\to 0}\large\frac{\sin x}{x}$
$\Rightarrow \lim\limits_{x\to 0}\bigg[\large\frac{x-\Large\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+.............}{x}\bigg]$
$\Rightarrow \lim\limits_{x\to 0}\big[1-\large\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+.........\big]$
$\Rightarrow 1$
As $x\to 0\large\frac{\sin x}{x}$ has a tendency to move closer and closer to 1 from the values which are less than 1.Hence the corresponding fractional part would also have a tendency to move towards 1.
Hence (d) is the correct answer.
answered Dec 31, 2013 by sreemathi.v

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