First slide

Introduction to limits

Question

$If l_{1} = lim_{x \to 2^{+}} (x + [x]), l_{2} = lim_{x \to 2^{-}} (2 x - [x]) and l_{3} = lim_{x \to \frac{π}{2}} \frac{\cos x}{x - \frac{π}{2}} then$

Moderate

A

$l_{1} < l_{2} < l_{3}$
B

$l_{2} < l_{3} < l_{1}$
C

$l_{3} < l_{2} < l_{1}$
D

$l_{1} < l_{3} < l_{2}$

Solution

$\begin{array}{l} Definition of [x] and L hospital rule \\ l_{1} = lim_{x \to 2^{+}} x + 2 \\ = 2 + 2 = 4 \\ l_{2} = lim_{x \to 2^{-}} 2 x - [x] \\ \underset{x \to 2^{-}}{= lim} 2 x - 1 \\ = 4 - 1 = 3 \\ l_{3} = lim_{x \to \frac{π}{2}} \frac{\cos x}{x - \frac{π}{2}} \\ = - 1 \end{array}$

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