Question

lim x-1 (x^3-1/x^2-6x+5)​

Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 1} \dfrac{x^{3} -1}{x^{2} - 6x + 5}$

As we know that,

Put the value of x = 1 in equation, and check their form.

$\sf \implies \lim_{x \to 1} \dfrac{(1)^{3} -1}{(1)^{2}-6(1) + 5 }$

$\sf \implies \lim_{x \to 1} \dfrac{0}{0}$

As we can see that it is in the form of 0/0.

So, we can simply Factorizes the equation, we get.

$\sf \implies \lim_{x \to 1} \dfrac{x^{3} -1}{x^{2} -6x + 5}.$

Factorizes the equation into middle term split, we get.

⇒ x² - 6x + 5.

⇒ x² - 5x - x + 5.

⇒ x(x - 5) - 1(x - 5).

⇒ (x - 1)(x - 5).

$\sf \implies \lim_{x \to 1} \dfrac{x^{3}-1 }{(x - 1)(x - 5)}$

As we can write x³ - 1 as,

⇒ x³ - 1 = x³ - 1³.

It is in the form of a³ - b³.

Formula of a³ - b³,

⇒ a³ - b³ = (a - b)(a² + ab + b²).

⇒ (x³ - 1³) = (x - 1)(x² + x + 1).

Put the value in equation, we get.

$\sf \implies \lim_{x \to 1} \dfrac{(x - 1)(x^{2} + x + 1)}{(x - 1)(x - 5)}$

Put the value of x = 1 in equation, we get.

$\sf \implies \lim_{x \to 1} \dfrac{(x^{2} + x+1) }{(x - 5)}$

$\sf \implies \lim_{x \to 1} \dfrac{1^{2}+ 1+ 1 }{(1 - 5) }$

$\sf \implies \lim_{x \to 1} \dfrac{3}{-4}$

$\sf \implies \lim_{x \to 1} = \dfrac{-3}{4}$

Answer 2

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

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$\tt \:\lim_{x\to1} \: \dfrac{ {x}^{3} - 1}{ {x}^{2} - 6x + 5}$

☆ On substituting directly x = 1, we get indeterminant form.

☆ So evaluate this, we use the concept of factorization.

☆ Use of identity and splitting of middle terms.

☆ identity used is

$\tt \ \: :  ⟼ {x}^{3} - {y}^{3} = (x - y)( {x}^{2} + xy + {y}^{2} )$

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☆ Now, Consider

$\tt \:\lim_{x\to1} \: \dfrac{ {x}^{3} - 1}{ {x}^{2} - 6x + 5}$

$\tt \:  = \tt \:\lim_{x\to1} \: \dfrac{ {x}^{3} - {1}^{3} }{ {x}^{2} - 5x - x \: + \: 5}$

$\tt \: = \:\lim_{x\to1} \:\dfrac{(x - 1)( {x}^{2} + x \times 1 + {1}^{2}) }{x(x - 5) - 1(x - 5)}$

$\tt \: = \:\lim_{x\to1} \:\dfrac{ \cancel{( x - 1)}( {x}^{2} + x + 1) }{ \cancel{(x - 1)}(x - 5)}$

$\tt \ \: = \dfrac{1 + 1 + 1}{1 - 5}$

$\tt \:  = \: - \: \dfrac{3}{4}$