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Q. $ \lim_{x \to {4}} \frac {3- \sqrt {5+x}} {x-4} $

Solution:

$\lim_{x\to4} \frac{3-\sqrt{5+x} }{x-4}$
= $\lim_{x\to4} \frac{\left(3-\sqrt{5+x}\right)\left(3+\sqrt{5+x}\right)}{\left(x-4\right)\left(3+\sqrt{5} +x\right) }$
= $\lim_{x\to4} \frac{9 - 5 -x}{\left(x-4\right)\left(3+ \sqrt{3+x}\right)}$
= $\lim_{x\to4} \frac{4-x}{\left(x-4\right)\left(3+\sqrt{5+x} \right)} $
= $\frac{-1}{3 +\sqrt{5 +4}} = - \frac{1}{6}$