Math Problem Statement

lim x tends to 1 x^x - x/x-1-logx

Solution

We are tasked with finding the limit:

limx1(xxxx1logx)\lim_{x \to 1} \left( \frac{x^x - x}{x - 1} - \log x \right)

Let’s proceed step by step.

Step 1: Analyze xxx^x as x1x \to 1

Recall that: xx=exlogxx^x = e^{x \log x} At x=1x = 1: xx=e1log1=e0=1x^x = e^{1 \cdot \log 1} = e^0 = 1 Thus, xxx=11=0x^x - x = 1 - 1 = 0 when x=1x = 1, which suggests that the expression xxxx1\frac{x^x - x}{x - 1} might have an indeterminate form. We'll need to apply L'Hopital's Rule to this part.

Step 2: Apply L'Hopital’s Rule to xxxx1\frac{x^x - x}{x - 1}

To apply L'Hopital's Rule, we differentiate the numerator and denominator.

  1. Numerator derivative f(x)=xxxf(x) = x^x - x:

    • Derivative of xxx^x: ddx(xx)=xx(logx+1)\frac{d}{dx} (x^x) = x^x (\log x + 1)
    • Derivative of x-x is 1-1.

    So the derivative of the numerator is: ddx(xxx)=xx(logx+1)1\frac{d}{dx} (x^x - x) = x^x (\log x + 1) - 1

  2. Denominator derivative g(x)=x1g(x) = x - 1: The derivative of x1x - 1 is simply 1.

Therefore, applying L'Hopital's Rule: limx1xxxx1=limx1(xx(logx+1)1)\lim_{x \to 1} \frac{x^x - x}{x - 1} = \lim_{x \to 1} \left( x^x (\log x + 1) - 1 \right) At x=1x = 1, this becomes: 11(log1+1)1=1×(0+1)1=11=01^1 (\log 1 + 1) - 1 = 1 \times (0 + 1) - 1 = 1 - 1 = 0

Step 3: Handle the logarithmic term

Now, we need to handle the logx-\log x term in the limit. As x1x \to 1, logx0\log x \to 0.

Step 4: Combine the results

We found that both the terms xxxx1\frac{x^x - x}{x - 1} and logx\log x tend to 0 as x1x \to 1. Therefore, the overall limit is: limx1(xxxx1logx)=00=0\lim_{x \to 1} \left( \frac{x^x - x}{x - 1} - \log x \right) = 0 - 0 = 0

Final Answer:

limx1(xxxx1logx)=0\lim_{x \to 1} \left( \frac{x^x - x}{x - 1} - \log x \right) = 0

Would you like more details or have any questions?

Here are 5 related questions:

  1. How can we prove xxx^x is differentiable at all points?
  2. What are some other ways to compute limits involving indeterminate forms?
  3. How does L'Hopital’s Rule work for higher-order indeterminate forms?
  4. Can we apply Taylor series expansion around x=1x = 1 to approximate the limit?
  5. How does the behavior of logarithmic functions compare to power functions near singularities?

Tip: L'Hopital’s Rule is useful when both the numerator and denominator tend to 0 or infinity; always check if these conditions hold before applying.

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Math Problem Analysis

Mathematical Concepts

Limits
L'Hopital's Rule
Logarithmic Functions
Exponential Functions

Formulas

lim_{x \to 1} (x^x - x)/(x - 1)
x^x = e^{x \log x}
L'Hopital's Rule: \lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}

Theorems

L'Hopital's Rule
Exponential Function Properties
Logarithmic Function Properties

Suitable Grade Level

Grades 11-12