# Thread: a question on calculus -limits

1. ## a question on calculus -limits

solve

lim ((1+2x)^0.5-(3x)^0.5)/((3+x)^0.5-2((x)^0.5))
x→1

-Riyafa-

2. $\lim_{x\to 1}\frac{\sqrt{1+2x}-\sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}}=\lim_{x\to 1}\frac{1-x}{\sqrt{1+2x}+\sqrt{3x}}\cdot\frac{\sqrt{3+x}+2\s qrt{x}}{3-3x}=$

$=\frac{1}{3}\lim_{x\to 1}\frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+\sqrt{3x }}=\frac{1}{3}\cdot\frac{4}{2\sqrt{3}}=\frac{2\sqr t{3}}{9}$

3. Originally Posted by riyafa
solve

lim ((1+2x)^0.5-(3x)^0.5)/((3+x)^0.5-2((x)^0.5))
x→1

-Riyafa-
Rationalize both top and bottom

$
\lim_{x \to 1} \frac{\sqrt{1+2x}- \sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}}
$

$
=\lim_{x \to 1} \frac{\sqrt{1+2x}- \sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}} \cdot \frac{\sqrt{1+2x}+ \sqrt{3x} }{\sqrt{1+2x}+ \sqrt{3x} } \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{3+x}+2\sqrt{x} }$

$
=\lim_{x \to 1} \frac{\sqrt{1+2x}- \sqrt{3x}}{\sqrt{3+x}-2\sqrt{x}} \cdot \frac{\sqrt{1+2x}+ \sqrt{3x} }{\sqrt{3+x}+2\sqrt{x} } \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+ \sqrt{3x} }$

$
=\lim_{x \to 1} \frac{1-x}{3 - 3x} \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+ \sqrt{3x} }$

$
=\lim_{x \to 1} \frac{1}{3} \cdot \frac{\sqrt{3+x}+2\sqrt{x}}{\sqrt{1+2x}+ \sqrt{3x} } = \frac{2}{3\sqrt{3}}$