First slide

Introduction to limits

Question

$Let f (x) = \{\begin{array}{l} b x + c for x > 1 \\ 3 c x - 2 b + 1 for x < 1 \end{array} then a relation between b and c so that$ ${Lt}_{x \to 1} f (x) exists is$

Moderate

A

$3 b + 2 c + 1 = 0$
B

$3 b + 2 c - 1 = 0$
C

$3 b - 2 c - 1 = 0$
D

$3 b - 2 c + 1 = 0$

Solution

$\underset{x \to 1}{L t} f (x) e x i s t s \Rightarrow \underset{x \to 1^{+}}{L t} f (x) = \underset{x \to 1^{-}}{L t} f (x) \Rightarrow \underset{x \to 1^{+}}{L t} \{b x + c\} = \underset{x \to 1^{-}}{L t} \{3 c x - 2 b + 1\}$
$\Rightarrow b + c = 3 c - 2 b + 1 \Rightarrow 3 b - 2 c - 1 = 0$

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