Question

Question number-13. Prove. ​

Answer 1

Step-by-step explanation:

The given limit will attain 0/0 indeterminate form by direct substitution.

Inorder to solve this problem, firstly we need to get rid of the cube root in numerator. We are aware about the following algebraic identity:

• $\boxed{ \sf A^3 - B^3 = (A-B)(A^2+B^2+AB) }$

Therefore, in the given limit, multiply both numerator and denominator by $\sf (A^2+B^2+AB)$.

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Here,

• $\sf A = \sqrt[3]{(1+x)}$
• $\sf B = \sqrt[3]{(1-x)}$.

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$\sf \implies \lim \limits_{x \to 0}\dfrac{(\sqrt[3]{1+x} - \sqrt[3]{1-x})[(\sqrt[3]{1+x})^2 + (\sqrt[3]{1-x})^2 + ( \sqrt[3]{1+x})( \sqrt[3]{1-x})]}{x[(\sqrt[3]{1+x})^2 + (\sqrt[3]{1-x})^2 + ( \sqrt[3]{1+x})( \sqrt[3]{1-x})]}$

Now use the identity in numerator:

$\sf \implies \lim \limits_{x \to 0}\dfrac{(\sqrt[3]{1+x})^{3} - (\sqrt[3]{1-x})^3}{x[(\sqrt[3]{1+x})^2 + (\sqrt[3]{1-x})^2 + ( \sqrt[3]{1+x})( \sqrt[3]{1-x})]}$

$\sf \implies \lim \limits_{x \to 0}\dfrac{1 + x - 1 + x}{x[(\sqrt[3]{1+x})^2 + (\sqrt[3]{1-x})^2 + ( \sqrt[3]{1+x})( \sqrt[3]{1-x})]}$

$\sf \implies \lim \limits_{x \to 0}\dfrac{2 \not x}{ \not x[(\sqrt[3]{1+x})^2 + (\sqrt[3]{1-x})^2 + ( \sqrt[3]{1+x})( \sqrt[3]{1-x})]}$

$\sf \implies \lim \limits_{x \to 0}\dfrac{2 }{ (\sqrt[3]{1+x})^2 + (\sqrt[3]{1-x})^2 + ( \sqrt[3]{1+x})( \sqrt[3]{1-x})}$

Now substitute the limits

$\sf \implies \dfrac{2 }{ (\sqrt[3]{1+0})^2 + (\sqrt[3]{1-0})^2 + ( \sqrt[3]{1+0})( \sqrt[3]{1-0})}$

$\sf \implies \dfrac{2 }{1 + 1 + ( 1)(1)}$

$\sf \implies \dfrac{2 }{1 + 1 + 1}$

$\sf \implies \dfrac{2 }{3}$

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So our required result is proved that:

$\underline{ \boxed{ \tt\lim \limits_{x \to 0}\dfrac{(\sqrt[3]{1+x} - \sqrt[3]{1-x})}{x} = \frac{2}{3} }}$

$\rule{280}{1}$

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