Get the information you need with the help of IDNLearn.com's expert community. Discover the information you need from our experienced professionals who provide accurate and reliable answers to all your questions.
Sagot :
Let's evaluate the limit:
[tex]\[ \lim_{x \to 1} \frac{x^{1/4} - 1}{x^{1/3} - 1} \][/tex]
### Step-by-Step Solution:
1. Check the indeterminate form:
First, we substitute [tex]\( x = 1 \)[/tex] directly into the expression to see what form it takes:
[tex]\[ \frac{1^{1/4} - 1}{1^{1/3} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \][/tex]
The expression is of the indeterminate form [tex]\(\frac{0}{0}\)[/tex], so we need to apply L'Hôpital's Rule.
2. Apply L'Hôpital's Rule:
L'Hôpital's Rule states that if [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)}\)[/tex] gives an indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can evaluate the limit of the derivatives:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
In this case, [tex]\( f(x) = x^{1/4} - 1 \)[/tex] and [tex]\( g(x) = x^{1/3} - 1 \)[/tex]. We need to find the derivatives of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( x^{1/4} - 1 \right) = \frac{1}{4} x^{-3/4} \][/tex]
[tex]\[ g'(x) = \frac{d}{dx} \left( x^{1/3} - 1 \right) = \frac{1}{3} x^{-2/3} \][/tex]
3. Evaluate the limit of the derivatives:
Now we have:
[tex]\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{\frac{1}{4} x^{-3/4}}{\frac{1}{3} x^{-2/3}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{\frac{1}{4} x^{-3/4}}{\frac{1}{3} x^{-2/3}} = \frac{\frac{1}{4}}{\frac{1}{3}} \cdot \frac{x^{-3/4}}{x^{-2/3}} = \frac{1}{4} \cdot \frac{3}{1} \cdot x^{2/3} \cdot x^{3/4} \][/tex]
Combining the exponents:
[tex]\[ = \frac{3}{4} \cdot x^{\left(-\frac{3}{4} + \frac{2}{3}\right)} = \frac{3}{4} \cdot x^{-\frac{9}{12} + \frac{8}{12}} = \frac{3}{4} \cdot x^{-1/12} \][/tex]
4. Simplify at [tex]\( x \to 1 \)[/tex]:
When [tex]\( x \)[/tex] approaches 1, the term [tex]\( x^{-1/12} \)[/tex] approaches [tex]\( 1 \)[/tex].
So:
[tex]\[ \lim_{x \to 1} \frac{3}{4} \cdot x^{-1/12} = \frac{3}{4} \cdot 1 = \frac{3}{4} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to 1} \frac{x^{1/4} - 1}{x^{1/3} - 1} = \frac{3}{4} \][/tex]
[tex]\[ \lim_{x \to 1} \frac{x^{1/4} - 1}{x^{1/3} - 1} \][/tex]
### Step-by-Step Solution:
1. Check the indeterminate form:
First, we substitute [tex]\( x = 1 \)[/tex] directly into the expression to see what form it takes:
[tex]\[ \frac{1^{1/4} - 1}{1^{1/3} - 1} = \frac{1 - 1}{1 - 1} = \frac{0}{0} \][/tex]
The expression is of the indeterminate form [tex]\(\frac{0}{0}\)[/tex], so we need to apply L'Hôpital's Rule.
2. Apply L'Hôpital's Rule:
L'Hôpital's Rule states that if [tex]\(\lim_{x \to c} \frac{f(x)}{g(x)}\)[/tex] gives an indeterminate form [tex]\(\frac{0}{0}\)[/tex] or [tex]\(\frac{\infty}{\infty}\)[/tex], we can evaluate the limit of the derivatives:
[tex]\[ \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)} \][/tex]
In this case, [tex]\( f(x) = x^{1/4} - 1 \)[/tex] and [tex]\( g(x) = x^{1/3} - 1 \)[/tex]. We need to find the derivatives of [tex]\( f(x) \)[/tex] and [tex]\( g(x) \)[/tex]:
[tex]\[ f'(x) = \frac{d}{dx} \left( x^{1/4} - 1 \right) = \frac{1}{4} x^{-3/4} \][/tex]
[tex]\[ g'(x) = \frac{d}{dx} \left( x^{1/3} - 1 \right) = \frac{1}{3} x^{-2/3} \][/tex]
3. Evaluate the limit of the derivatives:
Now we have:
[tex]\[ \lim_{x \to 1} \frac{f'(x)}{g'(x)} = \lim_{x \to 1} \frac{\frac{1}{4} x^{-3/4}}{\frac{1}{3} x^{-2/3}} \][/tex]
Simplify the fraction:
[tex]\[ \frac{\frac{1}{4} x^{-3/4}}{\frac{1}{3} x^{-2/3}} = \frac{\frac{1}{4}}{\frac{1}{3}} \cdot \frac{x^{-3/4}}{x^{-2/3}} = \frac{1}{4} \cdot \frac{3}{1} \cdot x^{2/3} \cdot x^{3/4} \][/tex]
Combining the exponents:
[tex]\[ = \frac{3}{4} \cdot x^{\left(-\frac{3}{4} + \frac{2}{3}\right)} = \frac{3}{4} \cdot x^{-\frac{9}{12} + \frac{8}{12}} = \frac{3}{4} \cdot x^{-1/12} \][/tex]
4. Simplify at [tex]\( x \to 1 \)[/tex]:
When [tex]\( x \)[/tex] approaches 1, the term [tex]\( x^{-1/12} \)[/tex] approaches [tex]\( 1 \)[/tex].
So:
[tex]\[ \lim_{x \to 1} \frac{3}{4} \cdot x^{-1/12} = \frac{3}{4} \cdot 1 = \frac{3}{4} \][/tex]
Therefore, the limit is:
[tex]\[ \lim_{x \to 1} \frac{x^{1/4} - 1}{x^{1/3} - 1} = \frac{3}{4} \][/tex]
Your participation means a lot to us. Keep sharing information and solutions. This community grows thanks to the amazing contributions from members like you. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.
