Question

Evaluate The Value of :

$\displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{1\ -\ \sqrt[3]{1\ +\ x^2}}}$

Don't Spam !!​

Answer 1

GIVEN :–

$\\ \implies \displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{1\ -\ \sqrt[3]{1\ +\ x^2}}}$

TO FIND :–

• Value of limit = ?

SOLUTION :–

• Let the function –

$\\ \sf \implies y = \displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{1\ -\ \sqrt[3]{1\ +\ x^2}}}$

• Now rationalization of denominator –

• We know that –

$\\ \sf \to {a}^{3} - {b}^{3} = (a - b)( {a}^{2} + ab + {b}^{2})$

$\\ \sf \to (a - b) = \dfrac{{a}^{3} - {b}^{3}}{( {a}^{2} + ab + {b}^{2})}$

• So that –

$\\ \sf \implies y = \displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2[ {(1)}^{2} + (1)(\sqrt[3]{1\ +\ x^2}) + ({\sqrt[3]{1\ +\ x^2}})^{2} ]}{(1)^{3} \ -\ (\sqrt[3]{1\ +\ x^2})^{3} }}$

$\\ \sf \implies y = \displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2[1 + (\sqrt[3]{1\ +\ x^2}) + ({\sqrt[3]{1\ +\ x^2}})^{2} ]}{1\ -\ (1\ +\ x^2)}}$

$\\ \sf \implies y = \displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2[1 + (\sqrt[3]{1\ +\ x^2}) + ({\sqrt[3]{1\ +\ x^2}})^{2} ]}{1\ -\ 1\ - \ x^2}}$

$\\ \sf \implies y = \displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{ \cancel{x^2}[1 + (\sqrt[3]{1\ +\ x^2}) + ({\sqrt[3]{1\ +\ x^2}})^{2} ]}{- \ \cancel{x^2}}}$

$\\ \sf \implies y = (-1) \displaystyle \sf{\lim_{x\ \to\ 0}[1 + (\sqrt[3]{1\ +\ x^2}) + ({\sqrt[3]{1\ +\ x^2}})^{2} ]}$

• Now Apply limits –

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

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

$\\ \sf \implies y = (-1) [1 + 1 + 1]$

$\\ \sf \implies y = (-1) [3]$

$\\ \large\implies { \boxed{ \sf y =-3}}$

Answer 2

Answer :

$-3$

Explanation :

We know, lim x => 0 ( 1 + x )^n ≈ ( 1 + nx )

=> $\displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{1\ -\ \sqrt[3]{1\ +\ x^2}}}$

=> $\displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{1\ -\ {1\ +\ x^2}}}1/3$

=> $\displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{1\ -\ {1\ +\ 1/3 \ x^2}}}$

=> $\displaystyle \sf{\lim_{x\ \to\ 0}\ \dfrac{x^2}{{1/3\ x^2}}}$

=> cancel ( $x^2$ )

=> $-1 / 3$

=> $-3$

‏‏‎ ‎

☆ Refer the attachment ⬆️

☆ ❥ So, It's Done ❥ !!

‏‏‎ ‎