First slide

Introduction to limits

Question

$f (x) = \frac{1 - x (1 + | 1 - x |)}{| 1 - x |} \cos (\frac{1}{1 - x})$ $x \neq 1$ . Then

Moderate

A

$lim_{x \to 1^{-}} f (x) = 0$ .
B

$lim_{x \to 1^{+}} f (x)$ does not exist
C

$lim_{x \to 1^{+}} f (x) = 0$
D

$lim_{x \to 1^{-}} f (x) does not exist$

Solution

We have,
$f (x) = \{\begin{cases} \frac{1 - x (1 + 1 - x)}{1 - x} \cos (\frac{1}{1 - x}) for x < 1 \\ \frac{1 - x (1 - (1 - x))}{- (1 - x)} \cos (\frac{1}{1 - x}) for x > 1 \end{cases}$
$\begin{array}{l} \Rightarrow f (x) = \{\begin{cases} \frac{1 - x (2 - x)}{1 - x} \cos (\frac{1}{1 - x}) & for x < 1 \\ \frac{1 - x^{2}}{- (1 - x)} \cos (\frac{1}{1 - x}) & for x > 1 \end{cases} \\ \Rightarrow f (x) = \{\begin{matrix} (1 - x) \cos (\frac{1}{1 - x}) for x < 1 \\ (- 1 - x) \cos (\frac{1}{1 - x}) for x > 1 \end{matrix} \\ ∴ lim_{x \to 1^{-}} f (x) = lim_{x \to 1} (1 - x) \cos (\frac{1}{1 - x}) \end{array}$
$= 0 \times$ (An oscillating number between -1 and 1)
$= 0$
$lim_{x \to 1^{+}} f (x) = lim_{x \to 1} - (1 + x) \cos (\frac{1}{1 - x})$
$= - 2$ (An oscillating number between -1 and 1)
$\Rightarrow lim_{x \to I^{+}} f (x)$ does not exist .

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