Let $[x]$ denote the greatest integer not exceeding $x$. If $l_1=\lim _{x \rightarrow 2^{+}}\left(x^2+[x]\right)$, $l_2=\lim _{x \rightarrow 3^{-}}(2 x-[x])$ and $l_3=\lim _{x \rightarrow \frac{\pi}{2}}\left(\frac{\cos x}{x-\frac{\pi}{2}}\right)$, then

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Aicharya · Accepted Answer

Given, $\begin{aligned} l_1 & =\lim _{x ightarrow 2^{+}}\left(x^{2}+[x]ight) \ & =\lim _{x ightarrow 2^{+}}\left(x^{2}+2ight) \quad\left[\because 	ext { as } x ightarrow 2^{+},[x]=2ight] \ & =4+2 \ l_1 & =6 \ l_2 & =\lim _{x ightarrow 3^{-}}(2 x-[x]) \quad\left[\because 	ext { as } x ightarrow 3^{-},[x]=2ight] \ & =\lim _{x ightarrow 3^{-}}(2 x-2) \quad \ & =2(3)-2 \ l_2 & =4 \ l_3 & =\lim _{x ightarrow \frac{\pi}{2}}\left(\frac{\cos x}{x-\frac{\pi}{2}}ight) \end{aligned}$ Put $x-\frac{\pi}{2}=y$ and as $x ightarrow \frac{\pi}{2}$, then $y ightarrow 0$ $\begin{aligned} x & =\frac{\pi}{2}+y \ & =\lim _{y ightarrow 0} \frac{\cos \left(\frac{\pi}{2}+yight)}{y} \ & =\lim _{y ightarrow 0}-\frac{\sin y}{y} \ l_3 & =-1 \ & l_3 < l_2 < l_1 \end{aligned}$ $	herefore$ Hence, answer is (d).

Let \([x]\) denote the greatest integer not exceeding \(x\). If \(l_1=\lim _{x \rightarrow…

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