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  • <img alt="\mathrm{\lim_{x\rightarrow 0^{+}}\frac{\sin \left [ x \right ]}{\left [ x \right ]}}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Clim_%7Bx%5Crightarrow%200%5E%7Ba

\mathrm{\lim_{x\rightarrow 0^{+}}\frac{\sin \left [ x \right ]}{\left [ x \right ]}}

Option: 1

0


Option: 2

1


Option: 3

2


Option: 4

5


Answers (1)

best_answer

In order for a right-hand limit \mathrm{\lim _{x \rightarrow a^{+}} f(x)} to make sense, there must exist a \mathrm{\delta>0} such that the function \mathrm{f} is defined in the open interval \mathrm{(a, a+\delta)}.
Since (with \mathrm{k} denoting an integer)

\mathrm{ f(x)=\frac{\sin [x]}{[x]}= \begin{cases}\frac{\sin k}{k} & k \leq x<k+1, k \neq 0, \\ \text { undefined } & 0 \leq x<1,\end{cases} }

the right-hand limit of \mathrm{ f \: at\: 0 } makes no sense.
(The dashed line in the plot is the graph \mathrm{ y=\sin x / x \: for \: x \neq 0 }.)

That said, it's conceivable the question (somewhat perversely) refers to the continuous extension
\mathrm{ g(x)= \begin{cases}\frac{\sin x}{x} & x \neq 0, \\ 1 & x=0,\end{cases} }
and that \mathrm{ f(x)=g([x]) }. If this interpretation were correct,

\mathrm{ f(x)=\frac{\sin [x]}{[x]}= \begin{cases}\frac{\sin k}{k} & k \leq x<k+1, k \neq 0, \\ 1 & 0 \leq x<1,\end{cases} }

\mathrm{ and \lim _{x \rightarrow 0^{+}} f(x)=1. }

Hence option 1 is correct.

 

Posted by

Nehul

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