ntaraxq

2022-07-07

Which rule or method can be applied against ?\

Karla Hull

Expert

$L=\underset{x\to 0}{lim}\frac{\sqrt{1+6x}-\left(1+3x\right)}{{x}^{2}}$
$=\underset{x\to 0}{lim}\frac{\left(\sqrt{1+6x}-\left(1+3x\right)\right)}{{x}^{2}}\frac{\left(\sqrt{1+6x}+\left(1+3x\right)\right)}{\left(\sqrt{1+6x}+\left(1+3x\right)\right)}$
$=\underset{x\to 0}{lim}\frac{\left(1+6x\right)-{\left(1+3x\right)}^{2}}{{x}^{2}\left(\sqrt{1+6x}+\left(1+3x\right)\right)}$
$=\underset{x\to 0}{lim}\frac{1+6x-\left(1+6x+9{x}^{2}\right)}{{x}^{2}\left(\sqrt{1+6x}+\left(1+3x\right)\right)}$
$=\underset{x\to 0}{lim}\frac{1+6x-1-6x-9{x}^{2}}{{x}^{2}\left(1+3x+\sqrt{1+6x}\right)}$
$=\underset{x\to 0}{lim}\frac{-9{x}^{2}}{{x}^{2}\left(1+3x+\sqrt{1+6x}\right)}$
$=\underset{x\to 0}{lim}\frac{-9}{\left(1+3x+\sqrt{1+6x}\right)}$
$=\underset{x\to 0}{lim}\frac{-9}{\left(1+1\right)}=-\frac{9}{2}$

auto23652im

Expert

$\underset{x\to 0}{lim}\frac{\frac{3}{\sqrt{1+6x}}-3}{2x}\to \frac{0}{0}$
So by L'Hopital's Rule,
$\underset{x\to 0}{lim}\frac{\frac{3}{\sqrt{1+6x}}-3}{2x}=\underset{x\to 0}{lim}\frac{-\frac{3}{2\left(1+6x{\right)}^{3/2}}\cdot 6}{2}$
$=\frac{\frac{-3}{2\cdot \left(1+6\left(0\right){\right)}^{3/2}}\cdot 6}{2}=-\frac{9}{2}$

Do you have a similar question?