Question

3. Calcule os seguintes limites (do tipo 0/0 envolvendo conjugado de radicais):

a) $\lim_{x \to 1}$ $\frac{ \sqrt{x} - 1}{x - 1}$


b) $\lim_{x \to 25}$ $\frac{x- 25}{ \sqrt{x} - 5}$

Answer 1

$\lim_{x \to 1} \frac{\sqrt{x}-1}{x-1}= \lim_{x \to 1} \frac{\sqrt{x}-1}{x-1}*\frac{\sqrt{x}+1}{\sqrt{x}+1}=\lim_{x \to 1}\frac{x-1}{(x-1)(\sqrt{x}+1)}=$
$\lim_{x \to 1}\frac{1}{\sqrt{x}+1}=\frac{1}{\sqrt{1}+1}=\boxed{\frac12}$

$\lim_{x \to 25} \frac{x-25}{\sqrt{x}-5}= \lim_{x \to 25} \frac{x-25}{\sqrt{x}-5}*\frac{\sqrt{x}+5}{\sqrt{x}+5}= \lim_{x \to 25}\frac{(x-25)(\sqrt{x}+5)}{x-25}=$
$\lim_{x \to 25}x-25=25-25=\boxed{0}$