First slide

Continuity

Question

$If f (x) = \{\begin{cases} \frac{| x + 2 |}{\tan_{}^{- 1} (x + 2)}, x \neq - 2 \\ 2, x = - 2 \end{cases} then f (x) is$

Moderate

A

$Continuous at x = - 2$
B

$not continuous at x = - 2$
C

$Differentiable at x = - 2$
D

$continuous but not differentiable at x = - 2$

Solution

$\begin{array}{l} \lim_{x \to - 2^{-}} f (x) = \lim_{x \to - 2^{-}} \frac{|x + 2|}{\tan^{- 1} (x + 2)} \\ = \lim_{x \to - 2^{-}} \frac{- (x + 2)}{\tan^{- 1} (x + 2)} = - 1 [∵ \lim_{x \to 0^{-}} \frac{\tan^{- 1} x}{x} = 1] \\ \lim_{x \to - 2^{+}} f (x) = \lim_{x \to - 2^{+}} \frac{|x + 2|}{\tan^{- 1} (x + 2)} \\ = \lim_{x \to - 2^{+}} \frac{x + 2}{\tan^{- 1} (x + 2)} = 1 \\ ∴ \lim_{x \to - 2} f (x) d o e s n o t e x i s t . \end{array}$

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