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

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SUMMARY

The limit of the complex fraction $\displaystyle\lim_{x \to 2} \dfrac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ results in an indeterminate form $\dfrac{0}{0}$ when evaluated directly. To resolve this, L'Hôpital's rule can be applied effectively. Alternatively, multiplying both the numerator and denominator by their respective conjugates is a viable method to simplify the expression and find the limit. Both approaches yield the same result, confirming the utility of L'Hôpital's rule in handling indeterminate forms.

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karush
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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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