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Question

If the Given limit: $lim x \to 4 \frac{4 x + 3}{x - 2}$ has value $\frac{k}{2} .$ Find the value of $k .$

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Solution

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$lim x \to 4 \frac{4 x + 3}{x - 2}$ $= \frac{4 (4) + 3}{4 - 2}$

$= \frac{16 + 3}{2}$

$= \frac{19}{2}$

Equating with $\frac{k}{2}$

we get, $k = 19$

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