Question

Assertion :If $\lambda$ and $\mu$ are positive real numbers and [*] denotes greatest integer function then ${}_{x\to 0^{+}}^{lim}\frac{x}{\lambda }\left[\frac{\mu }{x}\right]=\frac{\mu }{\lambda }$ Reason: ${}_{y\to \infty }^{lim}\frac{\left\{y\right\}}{y}=0$, where $\left\{\right\}$ denotes fractional part function.

• A. Assertion and Reason are correct and the Reason is the correct explanation for the Assertion.
• B. Assertion and Reason are correct and the Reason is the Not the correct explanation for the Assertion.
• C. Assertion is correct while the Reason is incorrect.
• D. Assertion is incorrect while the Reason is correct.

Answer 1

The correct option is A Assertion and Reason are correct and the Reason is the correct explanation for the Assertion.

$A:{lim}_{x\to 0^{+}}\frac{x}{\lambda }\left[\frac{\mu }{x}\right]$

$\frac{\mu }{x}-1\le \left[\frac{\mu }{x}\right]\le \frac{\mu }{x}$

$\frac{x}{\lambda }(\frac{\mu }{x}-1)\le \frac{x}{\lambda }\left[\frac{\mu }{x}\right]\le \frac{x}{\lambda }\times \frac{\mu }{x}$

${lim}_{x\to 0^{+}}(\frac{\mu }{\lambda }-\frac{x}{\lambda })\le {lim}_{x\to 0^{+}}\frac{x}{\lambda }\left[\frac{\mu }{x}\right]\le {lim}_{x\to 0^{+}}\frac{\mu }{\lambda }$

$\frac{\mu }{\lambda }-0\le {lim}_{x\to 0^{+}}\frac{x}{\lambda }\left[\frac{\mu }{x}\right]\le \frac{\mu }{\lambda }$

$\Rightarrow {lim}_{x\to 0^{+}}\frac{x}{\lambda }\left[\frac{\mu }{x}\right]=\frac{\mu }{\lambda }$

$R:{lim}_{y\to \infty }\frac{\left\{y\right\}}{y}={lim}_{y\to \infty }\frac{y-\left[y\right]}{y}$

$={lim}_{y\to \infty }(1-\frac{\left[y\right]}{y})$

$y-1\le \left[y\right]\le y$

$1-\frac{1}{y}\le \frac{\left[y\right]}{y}\le 1$

${lim}_{y\to \infty }(1-\frac{1}{y})\le {lim}_{y\to \infty }\frac{\left[y\right]}{y}\le {lim}_{y\to \infty }1$

$1\le {lim}_{y\to \infty }\frac{\left[y\right]}{y}\le 1$

$\Rightarrow 1\le {lim}_{1/y\to 0}\frac{\left[1/y\right]}{1/y}\le 1$