Question

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

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, we get.

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

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

As we can see that,

it is in the form of 0/0.

So we can factorizes the equation, we get.

⇒ x² - 6x + 5.

Factorizes into middle term split, we get.

⇒ x² - 5x - x + 5.

⇒ x(x - 5) - 1(x - 5).

⇒ (x - 1)(x - 5).

⇒ x³ - 1.

We can write as : 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 the equation, we get.

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

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

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

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

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

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

                                                                                                                       

MORE INFORMATION.

If function takes any of the following form, 0/0 & ∞/∞ then L'HOSPITAL'S RULE is applies.

$\sf \implies \lim_{x\to a} \dfrac{f(x)}{g(x)} = \sf \implies \lim_{x\to a} \dfrac{f'(x)}{g'(x)}$

NOTE :

L'HOSPITAL'S RULE can be repeated required number of times in the same question.

Answer 2

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

$\implies \sf \:\lim_{x\to1} \: \dfrac{ {1}^{3} - 1}{ {x}^{2} - 6(1) + 5}$

$\implies \sf \:\lim_{x\to1} \: \dfrac{0}{0}$

By factorising the equation.

We get,

$\implies \sf x^2 - 6x + 5$

$\implies \sf x^2 - 5x - x + 5$

$\implies \sf x(x-5) - 1(x-5)$

$\implies \sf (x-1)(x-5)$

$\implies \sf x^3 - 1$

Now, using a³ - b³.

$\implies \sf a^3 - b^3 = (a-b)(a^2 + ab + b^2)$

$\implies \sf x^3 - 1^3 = (x-1)(x^2 + x + 1)$

Now, putting the value in the equation.

We get,

$\implies \sf \:\lim_{x\to1} \: \dfrac{(x-1)(x^2 + x + 1)}{(x-1)x-5)}$

$\implies \sf \:\lim_{x\to1} \: \dfrac{(x^2 + x + 1)}{(x-5)}$

Now,

Putting the value of x = 1 in equation.

We get,

$\implies \sf \:\lim_{x\to1} \: \dfrac{(1)^2 + 1 +1}{(x-5)}$

$\implies \sf \:\lim_{x\to1} \: \dfrac{3}{-4}$

$\implies \bf \:\lim_{x\to1} \: = \dfrac{-4}{3}$