Let [.] denote the greatest integer function. | vedclass

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(B) Assertion $(A)$: We know that $[x] = x - \{x\}$,where $\{x\}$ is the fractional part of $x$ and $0 \leq \{x\} < 1$. $\lim_{x \rightarrow \infty} \

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$A$ is true,$R$ is true; $R$ is not the correct explanation of $A$

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(B) Assertion $(A)$: We know that $[x] = x - \{x\}$,where $\{x\}$ is the fractional part of $x$ and $0 \leq \{x\} $\lim_{x \rightarrow \infty} \frac{[x]}{x} = \lim_{x \rightarrow \infty} \frac{x - \{x\}}{x} = \lim_{x \rightarrow \infty} (1 - \frac{\{x\}}{x})$. Since $0 \leq \{x\} Thus,$\lim_{x \rightarrow \infty} \frac{[x]}{x} = 1 - 0 = 1$. So,$A$ is true. Reason $(R)$: Given $f(x) = x - 1$ and $h(x) = x$. $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{x - 1}{x} = \lim_{x \rightarrow \infty} (1 - \frac{1}{x}) = 1$. $\lim_{x \rightarrow \infty} \frac{h(x)}{x} = \lim_{x \rightarrow \infty} \frac{x}{x} = 1$. Both limits are $1$,so $R$ is true. However,$R$ does not explain why $\lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$.

Let $[.]$ denote the greatest integer function. Assertion $(A) : \lim_{x \rightarrow \infty} \frac{[x]}{x} = 1$. Reason $(R) : f(x) = x - 1, g(x) = [x], h(x) = x$ and $\lim_{x \rightarrow \infty} \frac{f(x)}{x} = \lim_{x \rightarrow \infty} \frac{h(x)}{x} = 1$.

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