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

Question

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, $I_2 < I_3 < I_1$, $I_1 < I_3 < I_2$, $I_1 < I_2 < I_3$, $I_3 < I_2 < I_1$

MARKS App · Accepted Answer

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

Search any question & find its solution

Looking for more such questions to practice?