Final answer to the problem
$-48$
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Step-by-step Solution
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- Solve using L'Hôpital's rule
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- Solve the limit using factorization
- Solve the limit using rationalization
- Integrate by partial fractions
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1
Factor the difference of cubes: $a^3-b^3 = (a-b)(a^2+ab+b^2)$
$\lim_{x\to2}\left(\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{6-x}-2}\right)$
Learn how to solve limits by rationalizing problems step by step online.
$\lim_{x\to2}\left(\frac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{6-x}-2}\right)$
Learn how to solve limits by rationalizing problems step by step online. Find the limit of (x^3-8)/((6-x)^(1/2)-2) as x approaches 2. Factor the difference of cubes: a^3-b^3 = (a-b)(a^2+ab+b^2). Applying rationalisation. Multiplying fractions \frac{\left(x-2\right)\left(x^2+2x+4\right)}{\sqrt{6-x}-2} \times \frac{\sqrt{6-x}+2}{\sqrt{6-x}+2}. Solve the product of difference of squares \left(\sqrt{6-x}-2\right)\left(\sqrt{6-x}+2\right).
Final answer to the problem
$-48$
