Question

prove that limit x tends to 1 x^3 -1/x^2-1=3/2​

Answer 1

EXPLANATION.

$\sf \implies \lim_{x \to 1}\dfrac{x^{3}- 1 }{x^{2} - 1} = \dfrac{3}{2}$

As we know that,

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

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

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

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

Now, we can see that it is a form of 0/0.

We can simply factorizes the equation, we get.

Formula of :

⇒ x³ - y³ = (x - y)(x² + xy + y²).

⇒ x² - y² = (x - y)(x + y).

Using the formula, we get.

⇒ (x³ - 1³) = (x - 1)(x² + x + 1).

⇒ (x² - 1²) = (x - 1)(x + 1).

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

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

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

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

$\sf \implies \lim_{x \to 1} \dfrac{3}{2}$

HENCE PROVED.

                                                                                                                                                 

MORE INFORMATION.

Factorization method.

If f(x) is of the form g(x)/h(x) and of indeterminant form then this form is removed by factorization g(x) and h(x) and cancel the common factors, then put the value of x.

Rationalization method.

In this method we rationalize the factor containing the square root and simplify and we put the value of x.

Answer 2

$\large\underline\blue{\bold{Given \:  Question :-  }}$

$\rm \:Prove \: that \: \displaystyle \rm \ \lim_{ x\to1} \dfrac{ {x}^{3} - 1 }{ {x}^{2} - 1} \: = \dfrac{3}{2}$

$\begin{gathered}\Large{\bold{\green{\underline{Identity \: \: used \::}}}} \end{gathered}$

$\boxed{ \pink{ \rm \:\:\lim_{x\to \: a}\dfrac{ {x}^{n} - {a}^{n} }{x - a} = {na}^{ n- 1} }}$

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

Consider, LHS,

We have,

$\rm : \implies \:\displaystyle \rm \ \lim_{ x\to1} \dfrac{ {x}^{3} - 1 }{ {x}^{2} - 1}$

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

So, to evaluate this limit,

• Divide both terms by x - 1, we get

$\rm : \implies \: \:\lim_{x\to \: 1} \: \dfrac{\dfrac{ {x}^{3} - 1}{x - 1} }{\dfrac{ {x}^{2} - 1 }{x - 1} }$

$\rm : \implies \:\:\dfrac{3 \times {(1)}^{3 - 1} }{2 \times {(1)}^{2 - 1} }$

$\rm : \implies \:\dfrac{3 \times {1}^{2} }{2 \times 1}$

$\rm : \implies \:\dfrac{3}{2}$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

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