First slide

Introduction to limits

Question

For what values of m does the $lim_{x \to 2} f (x)$ exist, when $f (x) = \{\begin{cases} mx - 3, when x < 2 \\ \frac{x}{m}, when x \geq 2 \end{cases}$

Easy

A

$\frac{1}{2}, 1$
B

$- \frac{1}{2}, 1$
C

-1,1
D

None of these

Solution

$lim_{x \to 2^{2}} f (x) = lim_{x \to 2^{-}} (mx - 3) = 2 m - 3$
$lim_{x \to 2^{+}} f (x) = lim_{x \to 2^{+}} \frac{x}{m} = \frac{2}{m}$
$lim_{x \to 2} f (x)$ exists when $lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x)$
$\begin{array}{lrl} \Rightarrow 2 m - 3 = \frac{2}{m} \Rightarrow 2 m^{2} - 3 m - 2 = 0 \\ ∴ m = - \frac{1}{2}, 2 \end{array}$

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