Limit with a lot of square roots

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Discussion Overview

The discussion revolves around evaluating the limit of the expression $$\lim_{x\rightarrow 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$$. Participants explore various methods to solve this limit, including algebraic manipulation and the application of L'Hospital's rule.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant suggests multiplying by the conjugate of the denominator to simplify the limit expression.
  • Another participant proposes using L'Hospital's rule to evaluate the limit, arriving at a specific value.
  • Some participants express a desire to find a solution without using L'Hospital's rule, indicating that their reference material has not covered it yet.
  • Alternative approaches are discussed, including approximating the square root for small values and substituting variables to simplify the limit.
  • A detailed algebraic manipulation is presented by a participant, showing the steps to rationalize both the numerator and denominator, ultimately leading to the same limit value as obtained using L'Hospital's rule.

Areas of Agreement / Disagreement

There is no consensus on a single method to solve the limit, as participants propose different approaches and express varying preferences for or against the use of L'Hospital's rule.

Contextual Notes

Participants mention different methods and their preferences, but the discussion does not resolve which method is superior or more appropriate given the context of their studies.

Ragnarok7
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I have the following problem:

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

If I multiply by the conjugate of the denominator I get

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

but am not sure where to go from here. Any suggestions? Thank you!
 
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Ragnarok said:
I have the following problem:

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

If I multiply by the conjugate of the denominator I get

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

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$\chi$ $\sigma$
 
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.
 
Ragnarok said:
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$\chi$ $\sigma$
 
Last edited:
Hello, Ragnarok!

$$\lim_{x\to 4}\frac{\sqrt{2x+1}-3}{\sqrt{x-2}-\sqrt{2}}$$
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)(<br /> \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}
 
Thank you so much! I wouldn't have thought to do it like this.
 

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