Easy Limits 2

Apply $\color{#3D99F6}{a^3-b^3=(a-b)(a^2+ab+b^2)}$ on numerator and then applying $\color{#3D99F6}{a^2-b^2=(a+b)(a-b)}$ .

$2x^{3}-128=2(\color{#D61F06}{x}^3-\color{#D61F06}{4}^3 )$ $=2(\color{#20A900}{\sqrt x}^2-\color{#20A900}{\sqrt 4}^2)(x^2+4x+16)$ $=2(\sqrt x+2)(\sqrt x-2)(x^2+4x+16)$ Limit simplifies to: $\Large\lim_{x\rightarrow 4}(2(\sqrt x+2)(x^2+4x+16))$ $\huge=\boxed{\color{#007fff}{384}}$

Substituting $x=4$ then limit takes the form $\frac{0}{0}$ .The derivative of the limit is $\dfrac{6x^2}{(\frac{1}{2} \times \frac{1}{\sqrt{x}})}$ . By L'Hopital's Rule , substituting the derivative of the limit with $x=4$ , we get the required limit as $96 \times 4 = \boxed{384}$ .

Review the Limit Property

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Observung the above graph we obtain the answer $384$ . $_\square$