Question

lim
$( \frac{ \sqrt{2x - 1 } \sqrt{x} - 1 }{ \sqrt{2x - 1} ?}$
x→1+

não consigo resolver esse limite​

Answer 1

Temos o seguinte limite

$\fbox{\displaystyle \lim_{x \to \ 1^+} \frac{\sqrt{2x-1}+\sqrt{x} - 1 }{\sqrt{2x-1}} }$

Então vamos substituir x = 1.

$\fbox{\displaystyle \lim_{x \to \ 1^+} \frac{\sqrt{2x-1}+\sqrt{x} - 1 }{\sqrt{2x-1}} \to \frac{\sqrt{2.1-1}+\sqrt{1} - 1 }{\sqrt{2.1-1}} }$

$\fbox{\displaystyle \frac{\sqrt{2-1}+ 1 - 1 }{\sqrt{2-1}} \to \frac{\sqrt{1} }{\sqrt{1}} = \frac{1}{1} = 1 }$

Portanto :

$\fbox{\displaystyle \lim_{x \to \ 1^+} \frac{\sqrt{2x-1}+\sqrt{x} - 1 }{\sqrt{2x-1}} = 1 }$