Math / Calculus

Questionما قيبة $\lim _{x \rightarrow-3} \frac{x^{3}+27}{x^{2}+4 x+3}=$
اخترأحد الذيارات $\frac{2}{27}$ $\frac{-27}{2}$ $\frac{27}{2}$ $\frac{-2}{27}$

Studdy Solution

STEP 1

1. We need to find the limit of the function as $x$ approaches $-3$ .
2. The function is given by $\frac{x^{3}+27}{x^{2}+4x+3}$ .
3. We will check for direct substitution and simplify if necessary.

STEP 2

1. Check if direct substitution is possible.
2. Factor the numerator and the denominator.
3. Simplify the expression by canceling common factors.
4. Substitute $x = -3$ into the simplified expression to find the limit.

STEP 3

Attempt direct substitution of $x = -3$ into the function:
$\frac{(-3)^{3}+27}{(-3)^{2}+4(-3)+3}$

STEP 4

Calculate the numerator and denominator separately:
Numerator: $(-3)^3 + 27 = -27 + 27 = 0$
Denominator: $(-3)^2 + 4(-3) + 3 = 9 - 12 + 3 = 0$
Since both the numerator and denominator are zero, direct substitution results in an indeterminate form $\frac{0}{0}$ .

STEP 5

Factor the numerator $x^3 + 27$ as a sum of cubes:
$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$
Factor the denominator $x^2 + 4x + 3$ :
$x^2 + 4x + 3 = (x + 3)(x + 1)$

STEP 6

Simplify the expression by canceling the common factor $(x + 3)$ :
$\frac{(x + 3)(x^2 - 3x + 9)}{(x + 3)(x + 1)} = \frac{x^2 - 3x + 9}{x + 1}$

STEP 7

Substitute $x = -3$ into the simplified expression:
$\frac{(-3)^2 - 3(-3) + 9}{-3 + 1} = \frac{9 + 9 + 9}{-2} = \frac{27}{-2}$
The value of the limit is:
$\boxed{\frac{-27}{2}}$

Was this helpful?

Questionما قيبة $\lim _{x \rightarrow-3} \frac{x^{3}+27}{x^{2}+4 x+3}=$
اخترأحد الذيارات $\frac{2}{27}$ $\frac{-27}{2}$ $\frac{27}{2}$ $\frac{-2}{27}$

Studdy Solution

STEP 1

1. We need to find the limit of the function as $x$ approaches $-3$ .
2. The function is given by $\frac{x^{3}+27}{x^{2}+4x+3}$ .
3. We will check for direct substitution and simplify if necessary.

STEP 2

1. Check if direct substitution is possible.
2. Factor the numerator and the denominator.
3. Simplify the expression by canceling common factors.
4. Substitute $x = -3$ into the simplified expression to find the limit.

STEP 3

Attempt direct substitution of $x = -3$ into the function:
$\frac{(-3)^{3}+27}{(-3)^{2}+4(-3)+3}$

STEP 4

Calculate the numerator and denominator separately:
Numerator: $(-3)^3 + 27 = -27 + 27 = 0$
Denominator: $(-3)^2 + 4(-3) + 3 = 9 - 12 + 3 = 0$
Since both the numerator and denominator are zero, direct substitution results in an indeterminate form $\frac{0}{0}$ .

STEP 5

Factor the numerator $x^3 + 27$ as a sum of cubes:
$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$
Factor the denominator $x^2 + 4x + 3$ :
$x^2 + 4x + 3 = (x + 3)(x + 1)$

STEP 6

Simplify the expression by canceling the common factor $(x + 3)$ :
$\frac{(x + 3)(x^2 - 3x + 9)}{(x + 3)(x + 1)} = \frac{x^2 - 3x + 9}{x + 1}$

STEP 7

Substitute $x = -3$ into the simplified expression:
$\frac{(-3)^2 - 3(-3) + 9}{-3 + 1} = \frac{9 + 9 + 9}{-2} = \frac{27}{-2}$
The value of the limit is:
$\boxed{\frac{-27}{2}}$

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Questionما قيبة lim⁡x→−3x3+27x2+4x+3=\lim _{x \rightarrow-3} \frac{x^{3}+27}{x^{2}+4 x+3}=x→−3lim​x2+4x+3x3+27​=اخترأحد الذيارات 227\frac{2}{27}272​ −272\frac{-27}{2}2−27​ 272\frac{27}{2}227​ −227\frac{-2}{27}27−2​

Studdy Solution

STEP 1

1. We need to find the limit of the function as x x x approaches −3-3−3.2. The function is given by x3+27x2+4x+3 \frac{x^{3}+27}{x^{2}+4x+3} x2+4x+3x3+27​.3. We will check for direct substitution and simplify if necessary.

STEP 2

1. Check if direct substitution is possible.2. Factor the numerator and the denominator.3. Simplify the expression by canceling common factors.4. Substitute x=−3 x = -3 x=−3 into the simplified expression to find the limit.

STEP 3

Attempt direct substitution of x=−3 x = -3 x=−3 into the function:(−3)3+27(−3)2+4(−3)+3 \frac{(-3)^{3}+27}{(-3)^{2}+4(-3)+3} (−3)2+4(−3)+3(−3)3+27​

STEP 4

STEP 5

Factor the numerator x3+27 x^3 + 27 x3+27 as a sum of cubes:x3+27=(x+3)(x2−3x+9) x^3 + 27 = (x + 3)(x^2 - 3x + 9) x3+27=(x+3)(x2−3x+9)Factor the denominator x2+4x+3 x^2 + 4x + 3 x2+4x+3:x2+4x+3=(x+3)(x+1) x^2 + 4x + 3 = (x + 3)(x + 1) x2+4x+3=(x+3)(x+1)

STEP 6

Simplify the expression by canceling the common factor (x+3) (x + 3) (x+3):(x+3)(x2−3x+9)(x+3)(x+1)=x2−3x+9x+1 \frac{(x + 3)(x^2 - 3x + 9)}{(x + 3)(x + 1)} = \frac{x^2 - 3x + 9}{x + 1} (x+3)(x+1)(x+3)(x2−3x+9)​=x+1x2−3x+9​

STEP 7

Substitute x=−3 x = -3 x=−3 into the simplified expression:(−3)2−3(−3)+9−3+1=9+9+9−2=27−2 \frac{(-3)^2 - 3(-3) + 9}{-3 + 1} = \frac{9 + 9 + 9}{-2} = \frac{27}{-2} −3+1(−3)2−3(−3)+9​=−29+9+9​=−227​ The value of the limit is:−272 \boxed{\frac{-27}{2}} 2−27​​

Get Studdy

Studdy solves anything!

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

Questionما قيبة $\lim _{x \rightarrow-3} \frac{x^{3}+27}{x^{2}+4 x+3}=$
اخترأحد الذيارات $\frac{2}{27}$ $\frac{-27}{2}$ $\frac{27}{2}$ $\frac{-2}{27}$

1. We need to find the limit of the function as $x$ approaches $-3$ .
2. The function is given by $\frac{x^{3}+27}{x^{2}+4x+3}$ .
3. We will check for direct substitution and simplify if necessary.

1. Check if direct substitution is possible.
2. Factor the numerator and the denominator.
3. Simplify the expression by canceling common factors.
4. Substitute $x = -3$ into the simplified expression to find the limit.

Attempt direct substitution of $x = -3$ into the function:
$\frac{(-3)^{3}+27}{(-3)^{2}+4(-3)+3}$

Factor the numerator $x^3 + 27$ as a sum of cubes:
$x^3 + 27 = (x + 3)(x^2 - 3x + 9)$
Factor the denominator $x^2 + 4x + 3$ :
$x^2 + 4x + 3 = (x + 3)(x + 1)$

Simplify the expression by canceling the common factor $(x + 3)$ :
$\frac{(x + 3)(x^2 - 3x + 9)}{(x + 3)(x + 1)} = \frac{x^2 - 3x + 9}{x + 1}$

Substitute $x = -3$ into the simplified expression:
$\frac{(-3)^2 - 3(-3) + 9}{-3 + 1} = \frac{9 + 9 + 9}{-2} = \frac{27}{-2}$
The value of the limit is:
$\boxed{\frac{-27}{2}}$