Limit with a lot of square roots

[Ragnarok7]

I have the following problem:

\(\displaystyle \lim_{x\rightarrow 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}\)

If I multiply by the conjugate of the denominator I get

\(\displaystyle \lim_{x\rightarrow 4}\frac{\sqrt{(2x+1)(x-2)}+\sqrt{2(2x+1)}-3\sqrt{x-2}-3\sqrt{2}}{x-4}\)

but am not sure where to go from here. Any suggestions? Thank you!

 

[chisigma]

I have the following problem:

\(\displaystyle \lim_{x\rightarrow 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}\)

If I multiply by the conjugate of the denominator I get

\(\displaystyle \lim_{x\rightarrow 4}\frac{\sqrt{(2x+1)(x-2)}+\sqrt{2(2x+1)}-3\sqrt{x-2}-3\sqrt{2}}{x-4}\)

but am not sure where to go from here. Any suggestions? Thank you!

In this case l'Hopital rule leads to the solution...

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

Kind regards$\chi$ $\sigma$

 

[Ragnarok7]

Thank you! Is it possible to solve this without L'Hospital's rule? The text I'm using hasn't introduced that in the section I got the problem from.

 

[chisigma]

Thank you! Is it possible to solve this without L'Hospital's rule? The text I'm using hasn't introduced that in the section I got the problem from.

All right!... an alternative is possible taking into account that 'for small t' is $\displaystyle \sqrt {1 + t} \sim 1 +\frac{t}{2}$ ...

Setting $x = 4 - \xi$ is ...

$\displaystyle \frac{\sqrt{2\ x +1}-3}{\sqrt{x-2}- \sqrt{2}}= \frac{\sqrt {9 -2\ \xi } - 3} {\sqrt{2 - \xi} - \sqrt{2}} \sim \frac{3}{\sqrt{2}}\ \frac{4}{9} = \frac{2\ \sqrt{2}}{3}\ (1)$

Kind regards$\chi$ $\sigma$

 

[soroban]

Hello, Ragnarok!

\(\displaystyle \lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}\)

Rationalize the denominator and the numerator.

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

$$\quad =\; \frac{\big([2x+1] - 9\big)( \sqrt{x-2} + \sqrt{2})}{\big([x-2]-2\big)(\sqrt{2x+1} + 3)}$$

$$\quad =\; \frac{(2x-8)(\sqrt{x-2} + \sqrt{2})}{(x-4)(\sqrt{2x+1} + 3)}$$

$$\quad =\; \frac{2(x-4)(\sqrt{x-2} + \sqrt{2})}{(x-4)(\sqrt{2x+1} + 3)}$$

$$\quad =\; \frac{2(\sqrt{x-2} + \sqrt{2})}{\sqrt{2x+1} + 3}$$$$\lim_{x\to4} \frac{2(\sqrt{x-2} + \sqrt{2})}{\sqrt{2x+1} + 3} \;=\;\frac{2(\sqrt{2}+\sqrt{2})}{\sqrt{9} + 3}$$

. . . . . $$=\;\frac{2(2\sqrt{2})}{6} \;=\;\frac{2\sqrt{2}}{3}$$

 

[Ragnarok7]

Thank you so much! I wouldn't have thought to do it like this.