Question

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Answer 1

28.

$</p><p>lim_{x \to 1}\frac{\sqrt{x}-1}{x-1}</p><p>$

$</p><p>=lim_{x \to 1}\frac{\sqrt{x}-1}{x-1}\times\frac{\sqrt{x}+1}{\sqrt{x}+1}</p><p>$

$</p><p>=lim_{x \to 1}\frac{\sqrt{x}^2-1^2}{(x-1)(\sqrt{x}+1)}</p><p>$

$</p><p>=lim_{x \to 1}\frac{x-1}{(x-1)(\sqrt{x}+1)}</p><p>$

$</p><p>=lim_{x \to 1}\frac{1}{\sqrt{x}+1}</p><p>$

$</p><p>=\frac{1}{1+1}=\frac{1}{2}</p><p>$

29.

$</p><p>lim_{x \to 9}\frac{3-\sqrt{x}}{9-x}</p><p>$

$</p><p>=lim_{x \to 9}\frac{3-\sqrt{x}}{9-x}\times\frac{3+\sqrt{x}}{3+\sqrt{x}}</p><p>$

$</p><p>=lim_{x \to 9}\frac{3^2-\sqrt{x}^2}{(9-x)(3+\sqrt{x})}</p><p>$

$</p><p>=lim_{x \to 9}\frac{9-x}{(9-x)(3+\sqrt{x})}</p><p>$

$</p><p>=lim_{x \to 9}\frac{1}{3+\sqrt{x}}</p><p>$

$</p><p>=\frac{1}{3+\sqrt{9}}=\frac{1}{6}</p><p>$

30.

$</p><p>lim_{x \to 64}\frac{x-64}{\sqrt{x}-8}</p><p>$

$</p><p>lim_{x \to 64}\frac{x-64}{\sqrt{x}-8}\times\frac{\sqrt{x}+8}{\sqrt{x}+8}</p><p>$

$</p><p>lim_{x \to 64}\frac{(x-64)(\sqrt{x}+8)}{\sqrt{x}^2-8^2}</p><p>$

$</p><p>lim_{x \to 64}\frac{(x-64)(\sqrt{x}+8)}{x-64}</p><p>$

$</p><p>lim_{x \to 64}\sqrt{x}+8=\sqrt{64}+8=16</p><p>$