Question

lim x tends to-1(x^3+1)/(x+1)​

Answer 1

EXPLANATION.

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

As we know that,

First we put the value of x = -1 in equation and check their indeterminant form.

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

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

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

As we can see,

This is the 0/0 indeterminant form,

In this form We can simply factorizes the equation.

Formula of :

⇒ x³ + y³ = (x + y)(x² - xy + y²).

We can write,

⇒ x³ + 1.

⇒ x³ + 1³ = (x + 1)(x² - x + 1²).

⇒ x³ + 1³ = (x + 1)(x² - x + 1).

Put the values in the equation, we get.

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

$\sf \implies \lim_{x \to -1} = x^{2} - x + 1.$

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

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

$\sf \implies \lim_{x \to -1} 1 + 1 + 1.$

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

                                                                                                                           

MORE INFORMATION.

Method of evaluation of limits.

(A) = When $\sf x \to \infty$

In this case expression should be expressed as a function 1/x and then after removing indeterminant form, (if it is there) replaced 1/x by 0.

(B) = Factorization method.

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

(C) = Rationalization method.

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

Answer 2

Answer:

$1 \frac{x {}^{3 } + 1} {x + 1}$

as we know that

first we put the value of x = -1 in equation and check their indeterminate form.

$- 1 \frac{( - 1) {}^{3} + 1}{( - 1) + 1}$

$1 \frac{ - 1 + 1}{ - 1 + 1}$

$- 1 \frac{0}{0}$

as we and see

This is the 0/0 indeterminate form , in this form we can simply factorizies the equation

Formula of :

${x}^{3} + y {}^{3} = (x + y)( {x}^{2} - xy + {y}^{2} )$

we can write.

${x}^{3} + 1$

${x}^{3} + {1}^{3} = (x + 1)( {x}^{2} - x + 1)$

put the value of the equation, we get.

$1 \frac{(x + 1)( {x}^{2} - x + 1)}{(x + 1)}$

$1 = {x}^{2} - x + 1$

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

$- 1 = ( - 1) {}^{2} - ( - 1) + 1$

$= 1 {}^{1} + 1 + 1$

$1 = 3$