Answer :
Certainly! Let's find the limit as \( x \) approaches 1 for the expression \(\frac{x^7 - 1}{x^{20} - 1}\):
[tex]\[ \lim_{x \to 1} \frac{x^7 - 1}{x^{20} - 1} \][/tex]
We'll work through this step-by-step.
### Step 1: Recognize Indeterminate Form
When we substitute \( x = 1 \) directly into the expression, both the numerator and denominator become zero:
[tex]\[ x^7 - 1 = 1^7 - 1 = 0 \][/tex]
[tex]\[ x^{20} - 1 = 1^{20} - 1 = 0 \][/tex]
Hence, we have a \(\frac{0}{0}\) indeterminate form, and we need to apply a method to resolve this.
### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have a \(\frac{0}{0}\) indeterminate form, we can take the derivatives of the numerator and the denominator:
[tex]\[ \lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{f'(x)}{g'(x)} \][/tex]
In this case:
[tex]\[ f(x) = x^7 - 1 \][/tex]
[tex]\[ g(x) = x^{20} - 1 \][/tex]
We'll differentiate both functions with respect to \( x \):
[tex]\[ f'(x) = \frac{d}{dx}(x^7 - 1) = 7x^6 \][/tex]
[tex]\[ g'(x) = \frac{d}{dx}(x^{20} - 1) = 20x^{19} \][/tex]
### Step 3: Substitute the Derivatives
Now we substitute the derivatives back into the limit:
[tex]\[ \lim_{x \to 1} \frac{7x^6}{20x^{19}} \][/tex]
### Step 4: Simplify the Expression
We can simplify this new expression:
[tex]\[ \lim_{x \to 1} \frac{7x^6}{20x^{19}} = \lim_{x \to 1} \frac{7}{20} x^{6-19} = \lim_{x \to 1} \frac{7}{20} x^{-13} \][/tex]
### Step 5: Evaluate the Limit
Now, we evaluate the limit as \( x \) approaches 1:
[tex]\[ \frac{7}{20} \cdot 1^{-13} = \frac{7}{20} \][/tex]
(Note that \( 1^{-13} = 1 \))
### Conclusion
So the limit is:
[tex]\[ \lim_{x \to 1} \frac{x^7 - 1}{x^{20} - 1} = \frac{7}{20} \][/tex]
And that's our final answer.
[tex]\[ \lim_{x \to 1} \frac{x^7 - 1}{x^{20} - 1} \][/tex]
We'll work through this step-by-step.
### Step 1: Recognize Indeterminate Form
When we substitute \( x = 1 \) directly into the expression, both the numerator and denominator become zero:
[tex]\[ x^7 - 1 = 1^7 - 1 = 0 \][/tex]
[tex]\[ x^{20} - 1 = 1^{20} - 1 = 0 \][/tex]
Hence, we have a \(\frac{0}{0}\) indeterminate form, and we need to apply a method to resolve this.
### Step 2: Apply L'Hôpital's Rule
L'Hôpital's Rule states that if we have a \(\frac{0}{0}\) indeterminate form, we can take the derivatives of the numerator and the denominator:
[tex]\[ \lim_{x \to 1} \frac{f(x)}{g(x)} = \lim_{x \to 1} \frac{f'(x)}{g'(x)} \][/tex]
In this case:
[tex]\[ f(x) = x^7 - 1 \][/tex]
[tex]\[ g(x) = x^{20} - 1 \][/tex]
We'll differentiate both functions with respect to \( x \):
[tex]\[ f'(x) = \frac{d}{dx}(x^7 - 1) = 7x^6 \][/tex]
[tex]\[ g'(x) = \frac{d}{dx}(x^{20} - 1) = 20x^{19} \][/tex]
### Step 3: Substitute the Derivatives
Now we substitute the derivatives back into the limit:
[tex]\[ \lim_{x \to 1} \frac{7x^6}{20x^{19}} \][/tex]
### Step 4: Simplify the Expression
We can simplify this new expression:
[tex]\[ \lim_{x \to 1} \frac{7x^6}{20x^{19}} = \lim_{x \to 1} \frac{7}{20} x^{6-19} = \lim_{x \to 1} \frac{7}{20} x^{-13} \][/tex]
### Step 5: Evaluate the Limit
Now, we evaluate the limit as \( x \) approaches 1:
[tex]\[ \frac{7}{20} \cdot 1^{-13} = \frac{7}{20} \][/tex]
(Note that \( 1^{-13} = 1 \))
### Conclusion
So the limit is:
[tex]\[ \lim_{x \to 1} \frac{x^7 - 1}{x^{20} - 1} = \frac{7}{20} \][/tex]
And that's our final answer.
