Solving limit without L'Hopital (can't finish)

Question

$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} $$

Im kind of a noob in this site, and it look like my questions miss some context, so im going to to the best I can to give it.

Here is what I tried and the farthest i got, I dont know how to continue:

Im sorry if this is not what context means.

Answer 1

write the term in the form $$\frac{(\sqrt{2x+1}-3)(\sqrt{2x+1}+3)(\sqrt{x-2}+\sqrt{2})}{(\sqrt{x-2}-\sqrt{2})(\sqrt{x-2}+\sqrt{2})(\sqrt{2x+1}+3)}$$

Answer 2

Let $\sqrt{2x+1}-3=u\implies2x=u^2+6u+8$

and $\sqrt{x-2}-\sqrt2=v\implies x= v^2+2\sqrt2v+4$

and as $x\to0; u,v\to0$

On division, $2v^2+4\sqrt2v=u^2+6u\iff v(2v+4\sqrt2)=u(u+6)$

$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}=\lim_{u,v\to0}\dfrac uv= \lim_{u,v\to0}\dfrac{2v+4\sqrt2}{u+6}=?$$

Answer 3

$$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}=$$

$$=\lim_{x\to 4}\frac{\frac{\sqrt{2x+1}-3}{x-4}}{\frac{\sqrt{x-2}-\sqrt{2}}{x-4}}=$$

$$=\frac{\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{x-4}}{\lim_{x\to 4}\frac{\sqrt{x-2}-\sqrt{2}}{x-4}}$$

If you can use derivatives, then $$=\frac{(\sqrt{2x+1})'\bigg|_{x=4}}{(\sqrt{x-2})'\bigg|_{x=4}}$$

Or you can calculate, e.g., $\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{x-4}$ by multiplying by $\frac{\sqrt{2x+1}+3}{\sqrt{2x+1}+3}$ and using the formula $a^2-b^2=(a-b)(a+b)$.

Answer 4

Hint use the definition of derivative at x=4 as follows $$\lim_{x\to 4} \frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}} = \lim_{x\to 4} \frac{\sqrt{2x+1}-3}{x-4} \cdot\lim_{x\to 4} \frac{x-4}{\sqrt{x-2}-\sqrt{2}}=(\sqrt{2x+1})' \frac{1}{(\sqrt{x-2})'}\Big|_{x=4} = \frac{1}{3}\frac{2\sqrt2}{1}$$