Question

$\displaystyle \lim_{x \to 4} \rm{ \dfrac{ {2x}^{3} - 128 }{ \sqrt{x} - 2}}$
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Answer 1

Answer:

$\huge \purple{384}$

Step-by-step explanation:

$\displaystyle \lim_{x \to 4} \rm{ \dfrac{ {2x}^{3} - 128 }{ \sqrt{x} - 2}} \times \frac{ \sqrt{x} + 2}{ \sqrt{x} + 2 } </p><p> \\ \\ = \displaystyle \lim_{x \to 4} \rm{ \dfrac{( {2x}^{3} - 128)( \sqrt{x} + 2)}{ x - 4}} \\ \\ use \: \: l \: hopital \: rule \\ \\ = </p><p>\displaystyle \lim_{x \to 4} \rm{ \dfrac{2(3 {x}^{2}) ( \sqrt{x} + 2) + \frac{1}{2 \sqrt{x}} (2 {x}^{3} - 128)} { 1}} \\ \\ = 6 \times {4}^{2} \times ( \sqrt{4} + 2) + \frac{1}{2 \sqrt{4} } \times (2. {4}^{3} - 128) \\ \\ = 6 \times 64 + \frac{1}{4} (128 - 128) \\ \\ = 384$

Answer 2

Answer:

Step-by-step explanation:

The answer is 384

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