First slide

Introduction to limits

Question

Let $f (x) = \{\begin{cases} x^{2}, x \in Z \\ \frac{k (x^{2} - 4)}{2 - x}, x \notin Z \end{cases}$
Then, $lim_{x \to 2} f (x)$

Moderate

A

exists only when $k = 1$
B

exists for every real $k$
C

exists for every real k except $k = 1$
D

does not exist

Solution

We have,
$\begin{array}{l} lim_{x \to 2^{-}} f (x) = lim_{h \to 0} f (2 - h) \\ \Rightarrow lim_{x \to 2^{-}} f (x) = lim_{h \to 0} \frac{k ((2 - h)^{2} - 4)}{2 - (2 - h)} = k lim_{h \to 0} \frac{h (h - 4)}{h} = - 4 k \end{array}$
$\begin{array}{l} lim_{x \to 2^{+}} f (x) = lim_{h \to 0} f (2 + h) = lim_{h \to 0} \frac{k ((2 + h)^{2} - 4)}{2 - (2 + h)} \\ \Rightarrow lim_{x \to 2^{+}} f (x) = k lim_{h \to 0} \frac{h (h + 4)}{- h} = - 4 k \\ ∴ lim_{x \to 2^{-}} f (x) = lim_{x \to 2^{+}} f (x) for all k \in R . \end{array}$

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