What is the limit as x approaches 0 of (\sqrt[3]{x^3+8}-sqrt(x^2+4))/(x^2) ?

The limit as x approaches 0 of (\sqrt[3]{x^3+8}-sqrt(x^2+4))/(x^2) is -1/4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

limit as x approaches 0 of (\sqrt[3]{x^3+8}-sqrt(x^2+4))/(x^2)

- \frac{1}{4}

Decimal

- 0.25

Solution steps

Apply L'Hopital's Rule

= x \to 0 lim \frac{\frac{x ^{2}}{( x ^{3} + 8 ) ^{\frac{2}{3}}} - \frac{x}{x ^{2} + 4}}{2 x}

Simplify

\frac{\frac{x ^{2}}{( x ^{3} + 8 ) ^{\frac{2}{3}}} - \frac{x}{x ^{2} + 4}}{2 x} : \frac{x x ^{2} + 4 - ( x ^{3} + 8 ) ^{\frac{2}{3}}}{2 ( x ^{3} + 8 ) ^{\frac{2}{3}} x ^{2} + 4}

= x \to 0 lim \frac{x x ^{2} + 4 - ( x ^{3} + 8 ) ^{\frac{2}{3}}}{2 ( x ^{3} + 8 ) ^{\frac{2}{3}} x ^{2} + 4}

Plug in the value

x = 0

= \frac{0 \cdot 0 ^{2} + 4 - ( 0 ^{3} + 8 ) ^{\frac{2}{3}}}{2 ( 0 ^{3} + 8 ) ^{\frac{2}{3}} 0 ^{2} + 4}

Simplify

\frac{0 \cdot 0 ^{2} + 4 - ( 0 ^{3} + 8 ) ^{\frac{2}{3}}}{2 ( 0 ^{3} + 8 ) ^{\frac{2}{3}} 0 ^{2} + 4} : - \frac{1}{4}

= - \frac{1}{4}

What is the limit as x approaches 0 of (\sqrt[3]{x^3+8}-sqrt(x^2+4))/(x^2) ?
The limit as x approaches 0 of (\sqrt[3]{x^3+8}-sqrt(x^2+4))/(x^2) is -1/4