Answer :
Sure, I'll walk you through the step-by-step solution to find the limit \(\lim _{x \rightarrow 3} \frac{x^3 - 27}{x - 3}\) using L'Hôpital’s rule.
1. Identify the Indeterminate Form:
First, we need to check whether the given limit results in an indeterminate form. Substitute \(x = 3\) into the expression:
[tex]\[ \frac{3^3 - 27}{3 - 3} = \frac{27 - 27}{0} = \frac{0}{0} \][/tex]
Since this is an indeterminate form of the type \(\frac{0}{0}\), we can apply L'Hôpital's rule.
2. Apply L'Hôpital's Rule:
L'Hôpital's rule states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then this limit can be evaluated as:
[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \][/tex]
where \(f'(x)\) and \(g'(x)\) are the derivatives of \(f(x)\) and \(g(x)\), respectively.
Here, \(f(x) = x^3 - 27\) and \(g(x) = x - 3\).
3. Differentiate the Numerator and Denominator:
Now, we calculate the derivatives of the numerator and the denominator.
- Derivative of the numerator \(f(x) = x^3 - 27\):
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 27) = 3x^2 \][/tex]
- Derivative of the denominator \(g(x) = x - 3\):
[tex]\[ g'(x) = \frac{d}{dx}(x - 3) = 1 \][/tex]
4. Substitute the Derivatives:
Substitute \(f'(x)\) and \(g'(x)\) back into the limit:
[tex]\[ \lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{x^3 - 27}{x - 3} = \lim_{x \to 3} \frac{3x^2}{1} \][/tex]
5. Evaluate the New Limit:
Finally, we evaluate the limit by substituting \(x = 3\):
[tex]\[ \lim_{x \to 3} \frac{3x^2}{1} = \frac{3(3)^2}{1} = \frac{3 \cdot 9}{1} = 27 \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to 3} \frac{x^3 - 27}{x - 3} = 27 \][/tex]
1. Identify the Indeterminate Form:
First, we need to check whether the given limit results in an indeterminate form. Substitute \(x = 3\) into the expression:
[tex]\[ \frac{3^3 - 27}{3 - 3} = \frac{27 - 27}{0} = \frac{0}{0} \][/tex]
Since this is an indeterminate form of the type \(\frac{0}{0}\), we can apply L'Hôpital's rule.
2. Apply L'Hôpital's Rule:
L'Hôpital's rule states that if \(\lim_{x \to a} \frac{f(x)}{g(x)}\) results in \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then this limit can be evaluated as:
[tex]\[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \][/tex]
where \(f'(x)\) and \(g'(x)\) are the derivatives of \(f(x)\) and \(g(x)\), respectively.
Here, \(f(x) = x^3 - 27\) and \(g(x) = x - 3\).
3. Differentiate the Numerator and Denominator:
Now, we calculate the derivatives of the numerator and the denominator.
- Derivative of the numerator \(f(x) = x^3 - 27\):
[tex]\[ f'(x) = \frac{d}{dx}(x^3 - 27) = 3x^2 \][/tex]
- Derivative of the denominator \(g(x) = x - 3\):
[tex]\[ g'(x) = \frac{d}{dx}(x - 3) = 1 \][/tex]
4. Substitute the Derivatives:
Substitute \(f'(x)\) and \(g'(x)\) back into the limit:
[tex]\[ \lim_{x \to 3} \frac{f(x)}{g(x)} = \lim_{x \to 3} \frac{x^3 - 27}{x - 3} = \lim_{x \to 3} \frac{3x^2}{1} \][/tex]
5. Evaluate the New Limit:
Finally, we evaluate the limit by substituting \(x = 3\):
[tex]\[ \lim_{x \to 3} \frac{3x^2}{1} = \frac{3(3)^2}{1} = \frac{3 \cdot 9}{1} = 27 \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to 3} \frac{x^3 - 27}{x - 3} = 27 \][/tex]
