Question

class 11th
find the limit of lim_x →3 x^4 - 81/2x^2 - 5x - 3

please answer ​

Answer 1

Answer

$\displaystyle \sf\lim_{x \to 3} \dfrac{x^4 - 81}{2x^2 - 5x -3} = \left. \dfrac{x^4 - 81}{2x^2 - 5x - 3} \right |_{x = 3}$

$\dfrac{(3)^4 - 81}{2(9)^2 - 5(3) - 3)} = \dfrac{0}{0}$

therefore, the function,

$\sf f(x) = \dfrac{ {x}^{4} - 81 }{2 {x}^{2} - 5x - 3 }$

$\sf f(x) = \dfrac{ ({x}^{2} - 9) - ( {x}^{2} - 9) }{2 {x}^{2} - 6x + x- 3 }$

$\sf f(x) = \dfrac{ (x + 3)(x - 3) ( {x}^{2} - 9) }{2x(x - 3) + (x- 3 )}$

$\sf f(x) = \dfrac{ (x + 3)(x - 3) ( {x}^{2} - 9) }{(2x + 1) (x- 3 )}$

$\sf f(x) = \dfrac{ (x + 3) ( {x}^{2} - 9) }{(2x + 1)}$

$\displaystyle\sf\lim_{x \to 3} f(x) = \left. \dfrac{ (x + 3) ( {x}^{2} - 9) }{2x(x - 3) } \right |_{x = 3}$

$\displaystyle\sf\lim_{x \to 3} f(x) = \dfrac{(3 + 3) \times (9 + 9) }{2(3) + 1 }$

$\displaystyle\sf\lim_{x \to 3} f(x) = \dfrac{6 \times 18}{7}$

$\underline {\boxed {\displaystyle\sf\lim_{x \to 3} f(x) = \dfrac{108}{7} }}$

Answer 2

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

$\rm \: \boxed{ \large\red{ {x}^{2} - {y}^{2} = (x + y)(x - y)}}$

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$\large\underline\purple{\bold{Solution :- }}$

$\tt \ \: :  ⟼ \tt \:\lim_{x\to3} \: \dfrac{ {x}^{4} - 81 }{ {2x}^{2} - 5x - 3}$

☆ On substituting directly x = 3, we get indeterminant form

☆ So, here we use the concept of factorization.

$\tt\implies \:\tt \:\lim_{x\to3}\dfrac{ {( {x}^{2}) }^{2} - {9}^{2} }{2 {x}^{2} - 6x + x - 3 }$

$\tt\implies \:\tt \:\lim_{x\to3}\dfrac{ ({x}^{2} - 9)( {x}^{2} + 9)}{2x(x - 3) + 1(x - 3)}$

$\tt\implies \:\tt \:\lim_{x\to3}\dfrac{ {x}^{2} - {3}^{2} }{(x - 3)(2x + 1)} \times \tt \:\lim_{x\to3}( {x}^{2} + 9)$

$\tt\implies \:18 \times \tt \:\lim_{x\to3}\dfrac{(x - 3)(x + 3)}{(x - 3)} \times \tt \:\lim_{x\to3}\dfrac{1}{2x + 1}$

$\tt\implies \:18 \times \dfrac{1}{7} \times \tt \:\lim_{x\to3}(x + 3)$

$\tt\implies \:\dfrac{18}{7} \times 6$

$\tt\implies \:\dfrac{108}{7}$