Question

lim x→-2 (x³+x²+4x+12)/(x³-3x+2)

Answer 1

Answer:

Answer of the Limit is 0

Step-by-step explanation:

In the limit question we should know the rule that limit approaches means that the limit is quietly near to that value but not that value so in this question limit is approaching to -2

so to find the value of the polynomial function we will plug in the -2 for the answer

Now given polynomial is

$\frac{x^{3}+x^{2}+4x+12}{x^{3}-3x+2}$

and

lim x → -2

Putting x = -2 in the given polynomial which is  

=$\frac{x^{3}+x^{2}+4x+12}{x^{3}-3x+2}$

=$\frac{(-2)^{3}+(-2)^{2}+4(-2)+12}{(-2)^{3}-3(-2)+2}$

=$\frac{-8+4-8+12}{-8+6+2}$

solving it gives

=$\frac{0}{0}$

We have got a $\frac{0}{0}$  form of the polynomial now we will solve this by L Hopital rule

this rule states that if limit is in this form then we make factors of the given polynomials and then solve them

So

Numerator of the given is

x³+x²+4x+12 = x²(x+1)+4(x+3)

It can not be further factorized

Denominator of the given is

x³-3x+2= x³-2x-x+2 = x²(x-2)-1(x-2)=(x-2)(x²-1)

Now the given polynomial becomes

$\frac{x^{2}(x+1)+4(x+3)}{(x-2)(x^{2}-1)}$

Putting x = -2

Gives us

= $\frac{(-2)^{2}(-2+1)+4(-2+3)}{(-2-2)((-2)^{2}-1)}$

= $\frac{4(-1)+4(1)}{(-4)(3)}$

= $\frac{4(-1)+4(1)}{(-4)(3)}$

= $\frac{0}{-12}$

=0

So Answer of the limit is 0