MHB What is the limit of a complex fraction with L'Hopital's rule?

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The limit of the complex fraction as x approaches 2 results in an indeterminate form of 0/0, prompting the use of L'Hôpital's rule or the conjugate method for evaluation. Participants discuss the effectiveness of L'Hôpital's rule in resolving the limit, while also considering the conjugate multiplication approach. Both methods are viable, with L'Hôpital's rule being favored for its straightforward application. The discussion emphasizes the importance of recognizing indeterminate forms and selecting appropriate techniques for limit evaluation. Ultimately, the limit can be resolved effectively using either method.
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\tiny{s8.1.6.64}
Evaluate
$\displaystyle\lim_{x \to 2}
\dfrac{\sqrt{6-x}{-2}}{\sqrt{3-x}-1}$

ok so if you plug in 2 directly you get $\dfrac{0}{0}$

So we either use L'H rule or use conjugate

or is there better way
 
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L'Hopital works

you can also multiply numerator and denominator by both conjugates
 
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