If [x] is the greatest integer function,then | vedclass

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(D) Let $x = 3 - h$,where $h \rightarrow 0^{+}$.As $x \rightarrow 3^{-}$,$|x| = x = 3 - h$ and $|3-x| = |3-(3-h)| = |h| = h$.Also,$[3x-9] = [3(3-h)-9]

Vedclass · Accepted Answer

$-2$

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(D) Let $x = 3 - h$,where $h \rightarrow 0^{+}$.As $x \rightarrow 3^{-}$,$|x| = x = 3 - h$ and $|3-x| = |3-(3-h)| = |h| = h$.Also,$[3x-9] = [3(3-h)-9] = [9-3h-9] = [-3h]$. Since $h > 0$,$-3h$ is a small negative number,so $[-3h] = -1$.And $[9-3x] = [9-3(3-h)] = [9-9+3h] = [3h]$. Since $h > 0$,$3h$ is a small positive number,so $[3h] = 0$.Substituting these into the limit expression:$\lim _{h \rightarrow 0^{+}} \frac{(3-(3-h)+\sin h) \cos(0)}{h(-1)} = \lim _{h \rightarrow 0^{+}} \frac{(h+\sin h)(1)}{-h} = \lim _{h \rightarrow 0^{+}} -\left(1 + \frac{\sin h}{h}\right) = -(1+1) = -2$.

If $[x]$ is the greatest integer function,then $\lim _{x \rightarrow 3^{-}} \frac{(3-|x|+\sin |3-x|) \cos [9-3 x]}{|3-x|[3 x-9]} = $

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Assertion $(A)$: $\lim _{x \rightarrow 0} \frac{1}{x} = \infty$
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