Answer :
Sure, let's find the limit [tex]\(\lim _{x \rightarrow 2}\left(\frac{x^3+7 x^2-36}{x^2+2 x-8}\right)\)[/tex] step-by-step.
1. Factorize the numerator and the denominator:
- Numerator: [tex]\(x^3 + 7x^2 - 36\)[/tex]
This can be factored by first checking if there is a common factor or using polynomial division. Here, it isn't immediately obvious, so we need a more involved factorization method.
- Denominator: [tex]\(x^2 + 2x - 8\)[/tex]
This quadratic expression can be factored directly:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) \][/tex]
2. Check for common factors:
Now we need to determine if the numerator [tex]\(x^3 + 7x^2 - 36\)[/tex] shares any common factors with the denominator [tex]\((x + 4)(x - 2)\)[/tex].
3. Substitute the value [tex]\(x = 2\)[/tex] into the factored denominator:
Notice when substituting [tex]\(x = 2\)[/tex] in the denominator:
[tex]\[ (x + 4)(x - 2) = (2 + 4)(2 - 2) = 6 \cdot 0 = 0 \][/tex]
This shows that directly substituting [tex]\(x = 2\)[/tex] makes the denominator zero, leading to an indeterminate form. Therefore, we need to determine if the numerator also becomes zero for [tex]\(x = 2\)[/tex].
4. Evaluate the numerator at [tex]\(x = 2\)[/tex]:
When [tex]\(x = 2\)[/tex]:
[tex]\[ 2^3 + 7 \cdot 2^2 - 36 = 8 + 28 - 36 = 0 \][/tex]
This shows that the numerator also equals zero at [tex]\(x = 2\)[/tex], confirming the [tex]\(0/0\)[/tex] indeterminate form.
5. Factorize the numerator to cancel the common term:
Since we have a [tex]\(0/0\)[/tex] form, we look to factorize the numerator to see if [tex]\((x - 2)\)[/tex] is a factor. After performing polynomial division, we find:
[tex]\[ x^3 + 7x^2 - 36 = (x - 2)(x^2 + 9x + 18) \][/tex]
6. Simplify the original expression:
The expression now becomes:
[tex]\[ \frac{(x - 2)(x^2 + 9x + 18)}{(x - 2)(x + 4)} \][/tex]
Cancel out the common [tex]\((x - 2)\)[/tex] term:
[tex]\[ \frac{x^2 + 9x + 18}{x + 4} \][/tex]
7. Calculate the limit with the simplified expression:
Substitute [tex]\(x = 2\)[/tex] into the simplified expression:
[tex]\[ \frac{2^2 + 9 \cdot 2 + 18}{2 + 4} = \frac{4 + 18 + 18}{6} = \frac{40}{6} = \frac{20}{3} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 2}\left(\frac{x^3+7x^2-36}{x^2+2x-8}\right) = \frac{20}{3} \][/tex]
1. Factorize the numerator and the denominator:
- Numerator: [tex]\(x^3 + 7x^2 - 36\)[/tex]
This can be factored by first checking if there is a common factor or using polynomial division. Here, it isn't immediately obvious, so we need a more involved factorization method.
- Denominator: [tex]\(x^2 + 2x - 8\)[/tex]
This quadratic expression can be factored directly:
[tex]\[ x^2 + 2x - 8 = (x + 4)(x - 2) \][/tex]
2. Check for common factors:
Now we need to determine if the numerator [tex]\(x^3 + 7x^2 - 36\)[/tex] shares any common factors with the denominator [tex]\((x + 4)(x - 2)\)[/tex].
3. Substitute the value [tex]\(x = 2\)[/tex] into the factored denominator:
Notice when substituting [tex]\(x = 2\)[/tex] in the denominator:
[tex]\[ (x + 4)(x - 2) = (2 + 4)(2 - 2) = 6 \cdot 0 = 0 \][/tex]
This shows that directly substituting [tex]\(x = 2\)[/tex] makes the denominator zero, leading to an indeterminate form. Therefore, we need to determine if the numerator also becomes zero for [tex]\(x = 2\)[/tex].
4. Evaluate the numerator at [tex]\(x = 2\)[/tex]:
When [tex]\(x = 2\)[/tex]:
[tex]\[ 2^3 + 7 \cdot 2^2 - 36 = 8 + 28 - 36 = 0 \][/tex]
This shows that the numerator also equals zero at [tex]\(x = 2\)[/tex], confirming the [tex]\(0/0\)[/tex] indeterminate form.
5. Factorize the numerator to cancel the common term:
Since we have a [tex]\(0/0\)[/tex] form, we look to factorize the numerator to see if [tex]\((x - 2)\)[/tex] is a factor. After performing polynomial division, we find:
[tex]\[ x^3 + 7x^2 - 36 = (x - 2)(x^2 + 9x + 18) \][/tex]
6. Simplify the original expression:
The expression now becomes:
[tex]\[ \frac{(x - 2)(x^2 + 9x + 18)}{(x - 2)(x + 4)} \][/tex]
Cancel out the common [tex]\((x - 2)\)[/tex] term:
[tex]\[ \frac{x^2 + 9x + 18}{x + 4} \][/tex]
7. Calculate the limit with the simplified expression:
Substitute [tex]\(x = 2\)[/tex] into the simplified expression:
[tex]\[ \frac{2^2 + 9 \cdot 2 + 18}{2 + 4} = \frac{4 + 18 + 18}{6} = \frac{40}{6} = \frac{20}{3} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \rightarrow 2}\left(\frac{x^3+7x^2-36}{x^2+2x-8}\right) = \frac{20}{3} \][/tex]
