If \(\rm\mathop {\lim }\limits_{x \to a} \frac{a^x -x^a}{x^x -a^a}= - 1\) , then what is the value of a?
1
Finding the Value of 'a' in a Limit Problem
The question asks us to find the value of 'a' given a specific limit equation:
\( \mathop {\lim }\limits_{x \to a} \frac{a^x -x^a}{x^x -a^a}= - 1 \)
Let's analyze the limit as \(x\) approaches \(a\). If we substitute \(x = a\) into the expression, we get:
- Numerator: \(a^a - a^a = 0\)
- Denominator: \(a^a - a^a = 0\)
This is an indeterminate form of type \(\frac{0}{0}\). Therefore, we can apply L'Hopital's Rule.
Applying L'Hopital's Rule
L'Hopital's Rule states that if \(\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)}\) is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\), then \(\mathop {\lim }\limits_{x \to c} \frac{f(x)}{g(x)} = \mathop {\lim }\limits_{x \to c} \frac{f'(x)}{g'(x)}\), provided the latter limit exists.
We need to find the derivatives of the numerator and the denominator with respect to \(x\).
Derivative of the Numerator
Let \(f(x) = a^x - x^a\). The derivative of \(a^x\) with respect to \(x\) is \(a^x \ln a\). The derivative of \(x^a\) with respect to \(x\) is \(a x^{a-1}\) (using the power rule \(\frac{d}{dx}(x^n) = nx^{n-1}\)).
So, \(f'(x) = \frac{d}{dx}(a^x - x^a) = a^x \ln a - a x^{a-1}\).
Derivative of the Denominator
Let \(g(x) = x^x - a^a\). The term \(a^a\) is a constant with respect to \(x\), so its derivative is 0. We need to find the derivative of \(x^x\). Let \(y = x^x\). We can use logarithmic differentiation:
- Take the natural logarithm of both sides: \(\ln y = \ln (x^x) = x \ln x\).
- Differentiate both sides with respect to \(x\): \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x \ln x)\).
- Using the product rule \(\frac{d}{dx}(uv) = u'v + uv'\): \(\frac{d}{dx}(x \ln x) = (1)(\ln x) + (x)(\frac{1}{x}) = \ln x + 1\).
- So, \(\frac{1}{y} \frac{dy}{dx} = \ln x + 1\).
- Substitute \(y = x^x\) back: \(\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)\).
So, the derivative of \(x^x\) is \(x^x(\ln x + 1)\).
Therefore, \(g'(x) = \frac{d}{dx}(x^x - a^a) = x^x(\ln x + 1) - 0 = x^x(\ln x + 1)\).
Evaluating the Limit after Applying L'Hopital's Rule
Now we can rewrite the limit using the derivatives:
\( \mathop {\lim }\limits_{x \to a} \frac{f'(x)}{g'(x)} = \mathop {\lim }\limits_{x \to a} \frac{a^x \ln a - a x^{a-1}}{x^x(\ln x + 1)} \)
Substitute \(x = a\) into this expression:
\( \frac{a^a \ln a - a (a)^{a-1}}{a^a(\ln a + 1)} \)
Simplify the term \(a \cdot a^{a-1}\): \(a^1 \cdot a^{a-1} = a^{1 + (a-1)} = a^a\).
So the expression becomes:
\( \frac{a^a \ln a - a^a}{a^a(\ln a + 1)} \)
Factor out \(a^a\) from the numerator:
\( \frac{a^a (\ln a - 1)}{a^a(\ln a + 1)} \)
Assuming \(a^a \neq 0\) (which is true for relevant values of 'a' for logarithms to be defined, i.e., \(a > 0\)), we can cancel \(a^a\):
\( \frac{\ln a - 1}{\ln a + 1} \)
Solving for 'a'
We are given that the original limit equals -1. Therefore, the evaluated limit must also equal -1:
\( \frac{\ln a - 1}{\ln a + 1} = -1 \)
Multiply both sides by \((\ln a + 1)\):
\( \ln a - 1 = -1 (\ln a + 1) \)
\( \ln a - 1 = -\ln a - 1 \)
Add \(\ln a\) to both sides:
\( \ln a + \ln a - 1 = -1 \)
\( 2 \ln a - 1 = -1 \)
Add 1 to both sides:
\( 2 \ln a = 0 \)
Divide by 2:
\( \ln a = 0 \)
To solve for \(a\), take the exponential of both sides (since \(\ln a = 0\) is equivalent to \(a = e^0\)):
\( a = e^0 \)
\( a = 1 \)
Let's check if \(a=1\) makes sense in the original limit definition (logarithms must be defined). \(\ln 1\) is defined, so \(a=1\) is a valid potential solution.
Thus, the value of 'a' that satisfies the given limit condition is 1.
| Step | Description | Result |
|---|---|---|
| 1 | Evaluate limit at \(x=a\) | \(\frac{0}{0}\) indeterminate form |
| 2 | Apply L'Hopital's Rule | Calculate derivatives of numerator and denominator |
| 3 | Numerator derivative \(\frac{d}{dx}(a^x - x^a)\) | \(a^x \ln a - a x^{a-1}\) |
| 4 | Denominator derivative \(\frac{d}{dx}(x^x - a^a)\) | \(x^x(\ln x + 1)\) |
| 5 | Evaluate \(\mathop {\lim }\limits_{x \to a} \frac{f'(x)}{g'(x)}\) | Substitute \(x=a\) into \(\frac{a^x \ln a - a x^{a-1}}{x^x(\ln x + 1)}\) |
| 6 | Simplify the expression | \(\frac{\ln a - 1}{\ln a + 1}\) |
| 7 | Set equal to -1 and solve for 'a' | \(\frac{\ln a - 1}{\ln a + 1} = -1 \implies \ln a = 0 \implies a = 1\) |
Revision Table: Key Concepts Used
| Concept | Brief Explanation |
|---|---|
| Limit Evaluation | Finding the value a function approaches as the input approaches some value. |
| Indeterminate Forms | Expressions like \(\frac{0}{0}\), \(\frac{\infty}{\infty}\), \(0 \cdot \infty\), \(\infty - \infty\), \(0^0\), \(1^\infty\), \(\infty^0\) that do not immediately reveal the limit value. |
| L'Hopital's Rule | A method to evaluate limits of indeterminate forms \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\) by taking the derivative of the numerator and denominator. |
| Derivative of \(a^x\) | \(\frac{d}{dx}(a^x) = a^x \ln a\) |
| Derivative of \(x^a\) | \(\frac{d}{dx}(x^a) = a x^{a-1}\) (Power Rule) |
| Derivative of \(x^x\) | \(\frac{d}{dx}(x^x) = x^x(\ln x + 1)\) (Using Logarithmic Differentiation) |
| Logarithmic Properties | \(\ln(e^k) = k\), \(e^{\ln k} = k\), \(\ln 1 = 0\), \(\ln x^n = n \ln x\). |
| Solving Equations with Logarithms | Using inverse functions (exponential) to isolate the variable. |
Additional Information: Logarithmic Differentiation
Logarithmic differentiation is a useful technique for finding the derivative of functions that are in the form of one function raised to the power of another function, like \(y = [f(x)]^{g(x)}\), or for complicated products and quotients.
The steps are:
- Take the natural logarithm of both sides: \(\ln y = \ln([f(x)]^{g(x)}) = g(x) \ln(f(x))\).
- Differentiate both sides with respect to \(x\): \(\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}[g(x) \ln(f(x))]\). Use the product rule and chain rule on the right side.
- Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = y \cdot \frac{d}{dx}[g(x) \ln(f(x))]\).
- Substitute the original expression for \(y\) back into the equation.
In this problem, we used it for \(x^x\):
- Let \(y = x^x\).
- \(\ln y = x \ln x\).
- \(\frac{1}{y} \frac{dy}{dx} = 1 \cdot \ln x + x \cdot \frac{1}{x} = \ln x + 1\).
- \(\frac{dy}{dx} = y (\ln x + 1) = x^x (\ln x + 1)\).
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