Find the values of the real constants $c$ and $d$ such that
$\lim_{x\to 0}\frac{\sqrt{c+dx} - \sqrt{3}}{x} = \sqrt{3} $
I really have no clue how to even get started.
Find the values of the real constants $c$ and $d$ such that
$\lim_{x\to 0}\frac{\sqrt{c+dx} - \sqrt{3}}{x} = \sqrt{3} $
I really have no clue how to even get started.
$\lim_{x\to 0}\frac{\sqrt{c+dx}-\sqrt{3}}{x}=\lim_{x\to 0}\frac{(\sqrt{c+dx}-\sqrt{3})(\sqrt{c+dx}+\sqrt{3})}{x(\sqrt{c+dx}+\sqrt{3})}=\lim_{x\to 0}\frac{c+dx-3}{x(\sqrt{c+dx}+\sqrt{3})}$. If this limit wants to be $\sqrt{3}$ so, we have to eliminate $x$ from the denominator. This makes $c=3$ and $d=6$. Check it.
In the comment to the other answer you ask for a method using L'Hôpital's rule. Note first that the only way the limit is going to exists is if $c = 3$. By L'Hôpital then you have $ \lim_{x \to 0}\frac{\sqrt{3 + dx} - \sqrt{3}}{x} = \lim_{x\to 0} \frac{d}{2\sqrt{3+dx}}. $
The only say that is going to equal $\sqrt{3}$ is if $d = 6$.