$\lim_{x\rightarrow 6} \frac{\sqrt{x+3}-3}{x-6}$ without L'Hôpital's rule

Question

I'm trying to do the following limit

$$\lim_{x\rightarrow 6} \frac{\sqrt{x+3}-3}{x-6}$$

without using L'Hôpital's rule.

Anyone knows any neat tricks that can be used?

Answer 1

Multiply by conjugate of numerator:

$$\frac{\sqrt{x+3}-3}{x-6}\cdot\frac{\sqrt{x+3}+3}{\sqrt{x+3}+3}=\frac{x-6}{(x-6)(\sqrt{x+3}+3)}=\frac1{\sqrt{x+3}+3}\xrightarrow[x\to6]{}\frac16$$

Answer 2

Rationalize:

$$\begin{align}\frac{\sqrt{x+3}-3}{x-6}&=\frac{\sqrt{x+3}-3}{x-6}\frac{\sqrt{x+3}+3}{\sqrt{x+3}+3}\\\\&=\frac{(\sqrt{x+3})^2-3^2}{(x-6)(\sqrt{x+3}+3)}\\\\&=\frac{\require{cancel}\cancel{x-6}}{\cancel{(x-6)}(\sqrt{x+3}+3)}\\\\&=\frac1{\sqrt{x+3}+3}\end{align}$$

Answer 3

$$\frac{\sqrt{x+3}-3}{x-6}=\frac{\sqrt{x+3}-3}{(x+3)-9}=\frac{\sqrt{x+3}-3}{(\sqrt{x+3}-3)(\sqrt{x+3}+3)}$$

Answer 4

As $u\to0$,

$$\sqrt{u+9}=3\sqrt{1+\frac{u}{9}}=3\left(1+\frac{u}{18}+o(u)\right)$$

Let $x=u+6$, then, as $u\to0$,

$$\frac{\sqrt{x+3}-3}{x-6}=\frac{\sqrt{u+9}-3}{u}=\frac{3+\frac{u}{6}-3+o(u)}{u}=\frac16+o(1)\to\frac16$$

Answer 5

Let $f(x)=\sqrt{x+3}$. Using taylor yields

$$f(x)=f(6)+f'(6)(x-6)+\mathcal O((x-6)^2)\\=3+\frac{1}{6}(x-6)+\mathcal O((x-6)^2)$$

So $\lim_{x\rightarrow 6} \frac{\sqrt{x+3}-3}{x-6}=\lim_{x\rightarrow 6}\frac{f(x)-3}{x-6}=\lim_{x\rightarrow 6}\frac{1}{6}+\mathcal O(x-6)=\frac{1}{6}+\lim_{x\rightarrow 6}\mathcal O(x-6)=\frac{1}{6}$

Answer 6

Am I the only one who thought of derivatives? This limit is exactly $f'(6)$ by definition!

$f'(x)=\frac{1}{2\sqrt{x+3}}$, so $f'(6)=\frac{1}{2\sqrt{9}}=\frac{1}{6}$.

EDIT: I saw now a comment about this, still surprised not a single answer did this, as this is not L'Hôpital.