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

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

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

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

3

Solution steps

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

Apply L'Hopital's Rule

: x \to 2 lim \frac{\frac{1}{2 x + 2}}{\frac{1}{3 ( x + 6 ) ^{\frac{2}{3}}}}

= x \to 2 lim \frac{\frac{1}{2 x + 2}}{\frac{1}{3 ( x + 6 ) ^{\frac{2}{3}}}}

Plug in the value

x = 2

= \frac{\frac{1}{2 2 + 2}}{\frac{1}{3 ( 2 + 6 ) ^{\frac{2}{3}}}}

Simplify

\frac{\frac{1}{2 2 + 2}}{\frac{1}{3 ( 2 + 6 ) ^{\frac{2}{3}}}} : 3

= 3

x \to \infty lim (x^{(\frac{l n ( 3 )}{1 + l n ( x )})})

x \to 0 lim (\frac{( 1 + x ) ^{x} - 1}{x})

x \to 1 - lim (x - 1)

x \to 9 lim (\frac{x ^{2} - 9}{x - 9})

h \to 0 lim (\frac{h ∣ h ∣}{h})

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