What is the limit as x approaches 2 of (\sqrt[3]{x^2+4}-\sqrt[4]{x^2+6x})/(x^2-4) ?

The limit as x approaches 2 of (\sqrt[3]{x^2+4}-\sqrt[4]{x^2+6x})/(x^2-4) is 1/192

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limit as x approaches 2 of (\sqrt[3]{x^2+4}-\sqrt[4]{x^2+6x})/(x^2-4)

x \to 2 lim (\frac{3 x ^{2} + 4 - 4 x ^{2} + 6 x}{x ^{2} - 4})

\frac{1}{192}

Decimal

0.00520 \dots

Solution steps

x \to 2 lim (\frac{3 x ^{2} + 4 - 4 x ^{2} + 6 x}{x ^{2} - 4})

Apply L'Hopital's Rule

: x \to 2 lim \frac{\frac{2 x}{3 ( x ^{2} + 4 ) ^{\frac{2}{3}}} - \frac{x + 3}{2 ( x ^{2} + 6 x ) ^{\frac{3}{4}}}}{2 x}

= x \to 2 lim \frac{\frac{2 x}{3 ( x ^{2} + 4 ) ^{\frac{2}{3}}} - \frac{x + 3}{2 ( x ^{2} + 6 x ) ^{\frac{3}{4}}}}{2 x}

Plug in the value

x = 2

= \frac{\frac{2 \cdot 2}{3 ( 2 ^{2} + 4 ) ^{\frac{2}{3}}} - \frac{2 + 3}{2 ( 2 ^{2} + 6 \cdot 2 ) ^{\frac{3}{4}}}}{2 \cdot 2}

Simplify

\frac{\frac{2 \cdot 2}{3 ( 2 ^{2} + 4 ) ^{\frac{2}{3}}} - \frac{2 + 3}{2 ( 2 ^{2} + 6 \cdot 2 ) ^{\frac{3}{4}}}}{2 \cdot 2} : \frac{1}{192}

= \frac{1}{192}

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What is the limit as x approaches 2 of (\sqrt[3]{x^2+4}-\sqrt[4]{x^2+6x})/(x^2-4) ?
The limit as x approaches 2 of (\sqrt[3]{x^2+4}-\sqrt[4]{x^2+6x})/(x^2-4) is 1/192